1111 比较麻烦的最短路

#include<cmath>

#include<map>

#include<iostream>

#include<cstring>

#include<cstdio>

#include<set>

#include<vector>

#include<queue>

#include<stack>

#include<algorithm>

using namespace std;

typedef long long ll;

const int N = 505;

const int INF = 0x3f3f3f3f;

#define MP(x, y) make_pair(x, y)

struct Node{

  int to, nx, di, ti;

}E[N*N*2];

int head[N], tot;

void add(int fr, int to, int di, int ti) {

  E[tot].to = to; E[tot].di = di; E[tot].ti = ti; E[tot].nx = head[fr];

  head[fr] = tot ++;

}

vector<int> ans1; int Ans1;

vector<int> ans2; int Ans2;

struct Hode{

  int po, di, ti;

  Hode(int a=0, int b=0, int c=0):po(a), di(b), ti(c){}

  bool operator < (const Hode & T) const {

    if(di != T.di) return di > T.di;

    else return ti > T.ti;

  }

};

struct Tode{

  int po, di, ti;

  Tode(int a=0, int b=0, int c=0):po(a), di(b), ti(c){}

  bool operator < (const Tode & T) const {

    if(ti != T.ti) return ti > T.ti;

    else return di > T.di;

  }

};

int pre[N], vis[N];

int dis[N], tim[N];

void dfs1(int x, int tag) {

  if(x == tag) return;

  dfs1(pre[x], tag);

  ans1.push_back(x);

}

void dfs2(int x, int tag) {

  if(x == tag) return;

  dfs2(pre[x], tag);

  ans2.push_back(x);

}

void dij1(int s, int t) {

  priority_queue<Hode> Q;

  memset(dis, INF, sizeof(dis));

  memset(tim, INF, sizeof(tim));

  dis[s] = 0; tim[s] = 0;

  memset(vis, 0, sizeof(vis));

  Q.push(Hode(s, dis[s], tim[s]));

  while(!Q.empty()) {

    int po = Q.top().po; Q.pop();

    if(vis[po]) continue;

    vis[po] =  1;

    for(int i = head[po]; ~i; i = E[i].nx) {

      int to = E[i].to;

      if(dis[to] > dis[po] + E[i].di) {

        dis[to] = dis[po] + E[i].di;

        tim[to] = tim[po] + E[i].ti;

        pre[to] = po;

        Q.push(Hode(to, dis[to], tim[to]));

      }else if(dis[to] == dis[po] + E[i].di && tim[to] > tim[po] + E[i].ti) {

        tim[to] = tim[po] + E[i].ti;

        pre[to] = po;

        Q.push(Hode(to, dis[to], tim[to]));

      }

    }

  }

//  printf("Distance = %d: %d", dis[t], s);

  ans1.push_back(s); Ans1 = dis[t];

  dfs1(t, s);

//  printf("\n");

}

void dij2(int s, int t) {

  priority_queue<Tode> Q;

  memset(dis, INF, sizeof(dis));

  memset(tim, INF, sizeof(tim));

  dis[s] = 0; tim[s] = 0;

  memset(vis, 0, sizeof(vis));

  Q.push(Tode(s, dis[s], tim[s]));

  while(!Q.empty()) {

    int po = Q.top().po; Q.pop();

    if(vis[po]) continue;

    vis[po] =  1;

    for(int i = head[po]; ~i; i = E[i].nx) {

      int to = E[i].to;

      if(tim[to] > tim[po] + E[i].ti) {

        dis[to] = dis[po] + 1;

        tim[to] = tim[po] + E[i].ti;

        pre[to] = po;

        Q.push(Tode(to, dis[to], tim[to]));

      }else if(tim[to] == tim[po] + E[i].ti && dis[to] > dis[po] + 1) {

        dis[to] = dis[po] + 1;

        pre[to] = po;

        Q.push(Tode(to, dis[to], tim[to]));

      }

    }

  }

  ans2.push_back(s); Ans2 = tim[t];

  dfs2(t, s);

}

int main() {

  int n, m;

  while(~scanf("%d %d", &n, &m)) {

    ans1.clear(); ans2.clear();

    memset(head, -1, sizeof(head));

    tot = 0;

    for(int i = 0; i < m; ++i) {

      int a, b, c, d, e; scanf("%d %d %d %d %d", &a, &b, &c, &d, &e);

      add(a, b, d, e);

      if(!c) add(b, a, d, e);

    }

    int s,t; scanf("%d %d", &s, &t);

    dij1(s, t);

    dij2(s, t);

  //  if(s == t) while(1);

    int fl = 1;

    for(int i = 0; i < min(ans1.size(), ans2.size()); ++i) {

      if(ans1[i] != ans2[i]) {

        fl = 0; break;

      }

    }

    if(fl) {

      printf("Distance = %d; Time = %d: ", Ans1, Ans2);

      for(int i = 0; i < ans1.size(); ++i) {

        if(i) printf(" -> ");

        printf("%d", ans1[i]);

      } printf("\n");

    }else {

      printf("Distance = %d: ", Ans1);

      for(int i = 0; i < ans1.size(); ++i) {

        if(i) printf(" -> ");

        printf("%d", ans1[i]);

      }

      printf("\nTime = %d: ", Ans2);

      for(int i = 0; i < ans2.size(); ++i) {

        if(i) printf(" -> ");

        printf("%d", ans2[i]);

      }

      printf("\n");

    }

  }

  return 0;

}

1112 gcd一下

#include<cmath>

#include<map>

#include<iostream>

#include<cstring>

#include<cstdio>

#include<set>

#include<vector>

#include<queue>

#include<stack>

#include<algorithm>

using namespace std;

typedef long long ll;

const int N = 505;

const int INF = 0x3f3f3f3f;

#define MP(x, y) make_pair(x, y)

char s[1005];

map<char, int> mp; int tot;

char table[40];

map<char, int> suc;

vector<int> vc[40];

int gcd(int a, int b) {

  if(b == 0) return a;

  else return gcd(b, a%b);

}

int main() {

  int n;

  tot = 0;

  for(char i = 'a'; i <= 'z'; ++i) mp[i] = ++tot, table[tot] = i;

  for(char i = '0'; i <= '9'; ++i) mp[i] = ++tot, table[tot] = i;

  mp['_'] = ++tot; table[tot] = '_';

  while(~scanf("%d", &n)) {

    suc.clear();

    for(int i = 0; i < 40; ++i) vc[i].clear();

    scanf("%s", s);

    int len = strlen(s);

    int tmp = 0;

    for(int i = 0; i <= len; ++i) {

      if(i && s[i] != s[i-1]) {

        vc[mp[s[i-1]]].push_back(tmp);

      //  printf("%c %d\n", s[i-1], tmp);

        tmp = 0;

      }

      tmp ++;

    }

    for(int i = 1; i <= tot; ++i) {

      if(vc[i].size() > 0) {

      //  printf("%c ", table[i]); for(int j = 0; i < vc[i].size(); ++j) printf("%d ", vc[i][j]); printf("\n");

        int tt = vc[i][0];

        for(int j = 1; j < vc[i].size(); ++j) {

          tt = gcd(tt, vc[i][j]);

        }

        if(tt % n == 0) {

      //    printf("%c\n",table[i]);

          suc[table[i]] = 1;

        }

      }

    }

    tmp = 0;

    for(int i = 0; i < len; ++i) {

      if(suc[s[i]] > 0) {

        suc[s[i]] = -1;

        printf("%c", s[i]);

      }

    }

    printf("\n");

    for(int i = 0; i <= len; ++i) {

      if(i && s[i] != s[i-1]) {

        if(suc[s[i-1]] == -1) {

          for(int j = 0; j < tmp/n; ++j) printf("%c", s[i-1]);

        }

        else for(int j = 0; j < tmp; ++j) printf("%c", s[i-1]);

        tmp = 0;

      }

      tmp ++;

    }

    printf("\n");

  }

  return 0;

}

1113

#include<cmath>

#include<map>

#include<iostream>

#include<cstring>

#include<cstdio>

#include<set>

#include<vector>

#include<queue>

#include<stack>

#include<algorithm>

using namespace std;

typedef long long ll;

const int N = 1e5+5;

const int INF = 0x3f3f3f3f;

#define MP(x, y) make_pair(x, y)

int A[N];

ll Abs(ll x) {

  if(x < 0) return -x;

  else return x;

}

int main() {

  int n;

  while(~scanf("%d", &n)) {

    ll all = 0;

    for(int i = 0; i < n; ++i) {

      scanf("%d", &A[i]);

      all += A[i];

    }

    sort(A, A+n);

    ll tmp = 0;

    for(int i = 0; i < n/2; ++i) {

      tmp += A[i];

    }

    ll ans = Abs(tmp - (all-tmp));

    if(n % 2) {

      tmp += A[n/2];

      ans = max(ans, Abs(tmp - (all-tmp)) );

    }

    printf("%d %lld\n", n%2, ans);

  }

  return 0;

}

1114 题意不好理解 其实就是并查集

#include<cmath>

#include<map>

#include<iostream>

#include<cstring>

#include<cstdio>

#include<set>

#include<vector>

#include<queue>

#include<stack>

#include<algorithm>

using namespace std;

typedef long long ll;

const int N = 1e5+5;

const int INF = 0x3f3f3f3f;

#define MP(x, y) make_pair(x, y)

int f[N];

int find(int x) { return x == f[x]? x : f[x] = find(f[x]); }

void merge(int a, int b) {

  if(b == -1 || a == -1) return;

  int t1 = find(a); int t2 = find(b);

  if(t1 != t2) {

    if(t1 > t2) f[t1] = t2; else f[t2] = t1;

  }

}

int has[N];

int sett[N], area[N];

int cnt[N]; double avg_set[N], avg_area[N];

int ord[N];

int cmp(int x, int y) {

  if(avg_area[x] != avg_area[y]) return avg_area[x] > avg_area[y];

  else return x < y;

}

int main() {

  int n;

  while(~scanf("%d", &n)) {

    for(int i = 0; i < 1e5; ++i) f[i] = i;

    for(int i = 0; i < n; ++i) {

      int a,b,c; scanf("%d %d %d", &a, &b, &c);

      has[a] ++; has[b] ++; has[c] ++;

      merge(a, b); merge(a, c);

      int d; scanf("%d", &d);

      for(int j = 0; j < d; ++j) {

        int l; scanf("%d", &l);

        has[l] ++;

        merge(a, l);

      }

      scanf("%d %d", &sett[a], &area[a]);

    }

//    printf("hh\n");

    for(int i = 0; i < 1e5; ++i) {

      if(has[i]) {

        int t1 = find(i);

        cnt[t1] ++; avg_set[t1] += sett[i]; avg_area[t1] += area[i];

      }

    }

    int tot = 0;

    for(int i = 0; i < 1e5; ++i) {

      if(cnt[i]) {

        avg_set[i] /= cnt[i]; avg_area[i] /= cnt[i];

        ord[tot ++] = i;

      }

    }

    sort(ord, ord + tot, cmp);

    printf("%d\n", tot);

    for(int i = 0; i < tot; ++i) {

      printf("%04d %d %.3f %.3f\n", ord[i], cnt[ord[i]], avg_set[ord[i]], avg_area[ord[i]]);

    }

  }

  return 0;

}

1115

#include<cmath>

#include<map>

#include<iostream>

#include<cstring>

#include<cstdio>

#include<set>

#include<vector>

#include<queue>

#include<stack>

#include<algorithm>

using namespace std;

typedef long long ll;

const int N = 1e3+5;

const int INF = 0x3f3f3f3f;

#define MP(x, y) make_pair(x, y)

int num[N];

int L[N];

int R[N];

int root = 0;

void insert(int x, int nw) {

  if(root == -1) {

    root = x; return;

  }

  if(num[x] <= num[nw]) {

    if(L[nw]) insert(x, L[nw]);

    else L[nw] = x;

  }else {

    if(R[nw]) insert(x, R[nw]);

    else R[nw] = x;

  }

}

int cnt[N];

int Deep;

void dfs(int x, int dep) {

  Deep = max(Deep, dep);

  cnt[dep] ++;

  if(L[x]) dfs(L[x], dep+1);

  if(R[x]) dfs(R[x], dep+1);

}

int main() {

  int n;

  while(~scanf("%d", &n)) {

    root = -1;

    for(int i = 0; i < n; ++i) {

      scanf("%d", &num[i]);

      insert(i, root);

    }

    Deep = -1;

    dfs(root, 0);

    printf("%d + %d = %d\n", cnt[Deep], cnt[Deep-1], cnt[Deep] + cnt[Deep-1]);

  }

  return 0;

}

1116

#include<cmath>

#include<map>

#include<iostream>

#include<cstring>

#include<cstdio>

#include<set>

#include<vector>

#include<queue>

#include<stack>

#include<algorithm>

using namespace std;

typedef long long ll;

const int N = 1e5+5;

const int INF = 0x3f3f3f3f;

#define MP(x, y) make_pair(x, y)

int Rank[N];

int has[N];

int prime(int x) {

  if(x == 1) return 0;

  if(x == 2) return 1;

  for(int i = 2; i <= sqrt(x); ++i) {

    if(x % i == 0)

      return 0;

  }

  return 1;

}

int main() {

  int n;

  while(~scanf("%d", &n)) {

    memset(has, 0, sizeof(has));

    for(int i = 0; i < n; ++i) {

      int a; scanf("%d", &a);

      Rank[a] = i+1;

    }

    int k; scanf("%d", &k);

    while(k--) {

      int a; scanf("%d", &a);

      if(has[a]) printf("%04d: Checked\n", a);

      else if(Rank[a] == 0) printf("%04d: Are you kidding?\n", a);

      else if(Rank[a] == 1) printf("%04d: Mystery Award\n", a);

      else if(prime(Rank[a])) printf("%04d: Minion\n", a);

      else printf("%04d: Chocolate\n", a);

      if(Rank[a]) has[a] = 1;

    }

  }

  return 0;

}

1117这题题意也是扯淡啊，，，，

#include<cmath>

#include<map>

#include<iostream>

#include<cstring>

#include<cstdio>

#include<set>

#include<vector>

#include<queue>

#include<stack>

#include<algorithm>

using namespace std;

typedef long long ll;

const int N = 1e5+5;

const int INF = 0x3f3f3f3f;

#define MP(x, y) make_pair(x, y)

int a[N];

int main() {

  int n;

  while(~scanf("%d", &n)) {

    for(int i = 0; i < n; ++i) scanf("%d", &a[i]);

    sort(a, a+n);

    a[n] = INF;

    int ans = -1;

    for(int i = 0; i < n; ++i) {

      if(a[i] != a[i+1]) {

        ans = max(ans, min(n-i-1, a[i+1]-1) );

      }

    }

    ans = max(ans, min(n, a[0]-1));

    printf("%d\n", ans);

  }

  return 0;

}

1118

include<cmath>

#include<map>

#include<iostream>

#include<cstring>

#include<cstdio>

#include<set>

#include<vector>

#include<queue>

#include<stack>

#include<algorithm>

using namespace std;

typedef long long ll;

const int N = 1e4+5;

const int INF = 0x3f3f3f3f;

#define MP(x, y) make_pair(x, y)

int fa[N];

int cnt[N];

int find(int x) { return x == fa[x]? x : fa[x] = find(fa[x]); }

int main() {

  int n;

  while(~scanf("%d", &n)) {

    memset(cnt, 0, sizeof(cnt));

    for(int i = 1; i <= 1e4; ++i) fa[i] = i;

    int m = -1;

    for(int i = 1; i <= n; ++i) {

      int a; scanf("%d", &a);

      int pre;

      if(a) {

        scanf("%d", &pre);

        m = max(m, pre);

      }

      for(int j = 1; j < a; ++j) {

        int b; scanf("%d", &b);

        m = max(m, b);

        int t1 = find(pre); int t2 = find(b);

        if(t1 != t2) fa[t2] = t1;

      }

    }

    for(int i = 1; i <= m; ++i) {

      cnt[find(i)] ++;

    }

    int all = 0; int maxx = -1;

    for(int i = 1; i <= m; ++i) {

      if(cnt[i]) {

        all ++;

        maxx = max(maxx, cnt[i]);

      }

    }

    printf("%d %d\n", all, m);

    int k; scanf("%d", &k);

    for(int i = 0; i < k; ++i) {

      int a; int b; scanf("%d %d", &a, &b);

      int t1 = find(a); int t2 = find(b);

      if(t1 == t2) printf("Yes\n");

      else printf("No\n");

    }

  }

  return 0;

}

1119 主要对于先后序的概念需要掌握

#include<cmath>

#include<map>

#include<iostream>

#include<cstring>

#include<cstdio>

#include<set>

#include<vector>

#include<queue>

#include<stack>

#include<algorithm>

using namespace std;

typedef long long ll;

const int N = 1e5+5;

const int INF = 0x3f3f3f3f;

#define MP(x, y) make_pair(x, y)

int pre[N];

int post[N];

int L[N], R[N];

int root;

int fl = 1;

void solve(int _l, int _r, int l, int r, int rt) {

//  printf("%d %d %d %d %d\n", _l,_r,l,r,rt);

  if(l > r) return;

  int ro = pre[_l];

  int pos = 0;

  for(int i = l; i <= r; ++i) {

    if(post[i] == ro) {

      pos = i; break;

    }

  }

  if(pos == l && pos != r) {

    L[rt] = post[pos];

    R[rt] = pre[_l+1];

    solve(_l+2, _r, l+1, r-1, pre[_l+1]);

  }else if(pos == r){

    fl = 0;

    L[rt] = post[pos];

    solve(_l+1, _r, l, r-1, post[pos]);

  }else {

  //  printf("hh %d\n", rt);

    L[rt] = post[pos];

    int _pos = (pos-l) + _l;

  //  printf("%d\n", _pos, pre[_pos]);

    R[rt] = pre[_pos+1];

    solve(_l+1, _pos, l, pos-1, post[pos]);

    solve(_pos+2, _r, pos+1, r-1, pre[_pos+1]);

  }

}

int nn = 0;

void dfs(int x) {

  if(L[x]) dfs(L[x]);

  if(!nn) nn = 1; else printf(" ");

  printf("%d", x);

  if(R[x]) dfs(R[x]);

}

int main() {

  int n;

  while(~scanf("%d", &n)) {

    memset(L, 0, sizeof(L));

    memset(R, 0, sizeof(R));

    nn = 0;

    for(int i = 1; i <= n; ++i) {

      scanf("%d", &pre[i]);

    }

    for(int i = 1; i <= n; ++i) {

      scanf("%d", &post[i]);

    }

    root = pre[1];

    solve(2, n, 1, n-1, root);

    if(fl) printf("Yes\n");

    else printf("No\n");

    dfs(root);

    printf("\n");

  }

  return 0;

}

1120

#include<cmath>

#include<map>

#include<iostream>

#include<cstring>

#include<cstdio>

#include<set>

#include<vector>

#include<queue>

#include<stack>

#include<algorithm>

using namespace std;

typedef long long ll;

const int N = 1e4+5;

const int INF = 0x3f3f3f3f;

#define MP(x, y) make_pair(x, y)

map<int, int> mp;

map<int, int>::iterator it;

void solve(int x) {

  int tt = 0;

  while(x) {

    tt += x%10;

    x /= 10;

  }

  mp[tt] ++;

}

int main() {

  int n;

  while(~scanf("%d", &n)) {

    mp.clear();

    for(int i = 0; i < n; ++i) {

      int a; scanf("%d", &a);

      solve(a);

    }

    int cnt = 0;

    for(it = mp.begin(); it != mp.end(); ++it) {

      if(it->second > 0) {

        cnt ++;

      }

    }

    printf("%d\n", cnt);

    int fl = 0;

    for(it = mp.begin(); it != mp.end(); ++it) {

      if(it->second > 0) {

        if(!fl) fl = 1; else printf(" ");

        printf("%d", it->first);

      }

    } printf("\n");

  }

  return 0;

}

pat1111-1120的更多相关文章

  1. CSU 1120 病毒（DP）

    题目链接:http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1120 解题报告:dp,用一个串去更新另一个串,递推方程是: if(b[i] > a ...

  2. AxureRP8实战手册（基础11-20）

    本文目录 基础11. 设置文本框输入为密码 基础12. 设置打开选择文件窗口 基础13. 限制文本框输入字符位数 基础14. 设置文本框提示文字 基础15. 设置文本框回车触发事件 基础16. 设置元 ...

  3. csuoj 1120: 病毒

    http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1120 1120: 病毒 Time Limit: 3 Sec  Memory Limit: 128 ...

  4. 九度OJ 1120 全排列 -- 实现C++STL中next_permutation()

    题目地址:http://ac.jobdu.com/problem.php?pid=1120 题目描述: 给定一个由不同的小写字母组成的字符串,输出这个字符串的所有全排列. 我们假设对于小写字母有'a' ...

  5. PAT 1120 Friend Numbers

    1120 Friend Numbers (20 分)   Two integers are called "friend numbers" if they share the sa ...

  6. 【Codeforces Round 1120】Technocup 2019 Final Round (Div. 1)

    Codeforces Round 1120 这场比赛做了\(A\).\(C\)两题,排名\(73\). \(A\)题其实过的有点莫名其妙...就是我感觉好像能找到一个反例(现在发现我的算法是对的... ...

  7. 日常英语---十三、MapleStory/Monsters/Level 11-20（邪恶之眼）

    日常英语---十三.MapleStory/Monsters/Level 11-20(邪恶之眼) 一.总结 一句话总结: evil ['ivl] A stronger version of Evil E ...

  8. 1120 Friend Numbers （20 分）

    1120 Friend Numbers (20 分) Two integers are called "friend numbers" if they share the same ...

  9. PAT甲级 1120. Friend Numbers (20)

    1120. Friend Numbers (20) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Two in ...

  10. PAT 1120 Friend Numbers[简单]

    1120 Friend Numbers (20 分) Two integers are called "friend numbers" if they share the same ...

随机推荐

  1. P1045 麦森数

    别问我为什么要写水题 #include <iostream> #include <cstdio> #include <cstring> #include <a ...

  2. CF 375D. Tree and Queries【莫队 | dsu on tree】

    题意: 一棵树,询问一个子树内出现次数$≥k$的颜色有几种 强制在线见上一道 用莫队不知道比分块高到哪里去了,超好写不用调7倍速度!!! 可以用分块维护出现次数这个权值,实现$O(1)-O(\sqrt ...

  3. BZOJ 1426: 收集邮票 [DP 期望 平方]

    传送门 题意: 有n种不同的邮票,皮皮想收集所有种类的邮票.唯一的收集方法是到同学凡凡那里购买,每次只能买一张,并且买到的邮票究竟是n种邮票中的哪一种是等概率的,概率均为1/n.但是由于凡凡也很喜欢邮 ...

  4. ES6，Array.includes()函数的用法

    在ES5,Array已经提供了indexOf用来查找某个元素的位置,如果不存在就返回-1,但是这个函数在判断数组是否包含某个元素时有两个小不足,第一个是它会返回-1和元素的位置来表示是否包含,在定位方 ...

  5. MySQL的字符编码设置

    -- 创建数据库时,设置数据库的编码方式 -- CHARACTER SET:指定数据库采用的字符集,utf8不能写成utf-8-- COLLATE:指定数据库字符集的排序规则,utf8的默认排序规则为 ...

  6. centos出现“FirewallD is not running”怎么办

    最近在阿里云服务器centos上安装了mysql数据库,默认是不开启远端访问功能,需要设置一下防火墙,在开放默认端口号 3306时提示FirewallD is not running,经过排查发现是防 ...

  7. 对 url 中含有的中文进行转码操作

    对 url 中含有的中文进行转码操作 一般情况下,将带有中文的 url 拷贝到开发工具,开发工具都会有相应的转码(自动转码), 现在大部分的浏览器也可以对含有中文的 url 进行转码(自动转码) 情景 ...

  8. Postfix+dovecot搭建简单邮箱服务器

    实验环境: (1)修改主机名:hostnamectl set-hostname mail.meilintong.com 退出,重新登陆 (2)关闭selinux (3)关闭防火墙 1.安装postfi ...

  9. WinForm中ClickOnce发布至广域网

    ClickOnce智能客户端,是微软提供比较早的一项技术,用于实现WinForm开发的应用程序能够自动更新,省去给每台客户端升级带来的困扰. 从网上的贴子里看,有的说好用,有的说不好用.客观的说,微软 ...

  10. 将Tomcat配置到你的mac电脑上，命令行启动tomcat

    1.下载tomcat7文件 2.编辑打开.bash_profile vim .bash_profile 3.在该文件最后面添加(CATALINA_HOME为tomcat解压目录) CATALINA_H ...