BUPT2017 springtraining(16) #6 ——图论

题目链接

A.容易发现最后字符的对应都是一对一的

或者说我们没办法出现最后多对一或者一对多的情况

所以只要算出 ‘a’ - 'z' 每个字符最后对应的字符即可

 #include <cstdio>
 #include <algorithm>

 int n, m, b[];

 char a[], s[];

 int main() {
     char s1[], s2[];
     scanf("%d %d %s", &n, &m, s);
     for(int i = ;i < ;i ++) a[i] = i + 'a', b[i] = i;
     while(m --) {
         scanf("%s %s", s1, s2);
         if(s1[] == s2[]) continue;
         a[b[s1[] - 'a']] = s2[];
         a[b[s2[] - 'a']] = s1[];
         std::swap(b[s1[] - 'a'], b[s2[] - 'a']);
     }
     for(int i = ;i < n;i ++)
         putchar(a[s[i] - 'a']);
     return ;
 }

B.

C.A simple directed graph is a directed graph having no multiple edges or graph loops.

graph loops其实说的是自环?道理我都懂，但是自环不是self loops之类的吗？

我们考虑连完所有边的最终形态，应该是

两个强连通分量AB，内部均为完全图

A中所有点都能到B中所有点

B中没有连向A中点的边

稍微算一下可知令siz_A与siz_B差距越大，可连边就越多

那我们先对原图求强连通分量

这里面没有入度或没有出度的scc是有可能成为A或B的scc

找出这里面siz最小的scc来确定A和B其中一个的siz即可

为什么不可能是siz最大的一个呢？

我们列一下式子发现时恒等式

siz最小的必然不比siz最大的结果差

#include <bits/stdc++.h>

using namespace std;

const int maxn = ;

int n, m, cnt, top, sum, c1[maxn], c2[maxn], siz[maxn];

int in[maxn], vis[maxn], sta[maxn], dfn[maxn], low[maxn], bel[maxn];

vector <int> e[maxn];

void tarjan(int x) {
    vis[x] = ;
    in[x] = , sta[++ top] = x, dfn[x] = low[x] = ++ cnt;
    for(int t, i = ;i < e[x].size();i ++) {
        t = e[x][i];
        if(!vis[t]) {
            tarjan(t);
            low[x] = min(low[x], low[t]);
        }
        else if(in[t] && low[t] < low[x]) low[x] = low[t];
    }
    if(low[x] == dfn[x]) {
        sum ++;
        for(int i = ;i != x;i = sta[top --], bel[i] = sum, in[i] = , siz[sum] ++);
    }
}

int main() {
    long long ans;
    int Case, u, v, tmp;
    ios::sync_with_stdio(false);
    cin >> Case;
    for(int tt = ;tt <= Case;tt ++) {
        cout << "Case " << tt << ": ";
        cin >> n >> m;
        cnt = ans = sum = , tmp = n;
        for(int i = ;i <= n;i ++) e[i].clear(), vis[i] = c1[i] = c2[i] = siz[i] = ;
        for(int i = ;i <= m;i ++) cin >> u >> v, e[u].push_back(v);
        for(int i = ;i <= n;i ++)
            if(!vis[i]) tarjan(i);
        if(sum == ) cout << - << endl;
        else {
            for(int i = ;i <= n;i ++)
                for(int j = ;j < e[i].size();j ++)
                    if(bel[i] != bel[e[i][j]]) c1[bel[i]] ++, c2[bel[e[i][j]]] ++;
            for(int i = ;i <= sum;i ++)
                if(c1[i] ==  || c2[i] == )
                    tmp = min(tmp, siz[i]);
             cout << 1ll * n * n - m - n - 1ll * tmp * (n - tmp) << endl;
        }
    }
}

D.其实只要先处理出昨天没走的人

就可以算出今天各个时刻的人数了，取max即可

 #include <iostream>

 using namespace std;

 bool in[];

 int n, ans, cnt, x;

 char str[][];

 int main() {
     ios::sync_with_stdio(false);
     cin >> n;
     for(int i = ;i <= n;i ++) {
         cin >> str[i] >> x;
         if(str[i][] == '+') in[x] = ;
         else if(!in[x]) cnt ++;
     }
     ans = cnt;
     for(int i = ;i <= n;i ++) {
         if(str[i][] == '+') cnt ++, ans = max(cnt, ans);
         else cnt --;
     }
     cout << ans;
 }

E.我们令第 i 个数为 a[i]

计算出以 a[i] 为序列第一个数的方案种数

以 a[i] 为序列第二个数的方案种数

ans += 以 a[i] / k 为序列第二个数的方案种数

 #include <map>
 #include <iostream>

 using namespace std;

 typedef long long ll;

 map <ll, ll> p[];

 int n, k;

 int a[];

 ll ans;

 int main() {
     ios::sync_with_stdio(false);
     cin >> n >> k;
     for(int i = ;i <= n;i ++) {
         cin >> a[i];
         if(a[i] % k == ) ans += p[][a[i] / k], p[][a[i]] += p[][a[i] / k];
         p[][a[i]] ++;
     }
     cout << ans;
     return ;
 }

F.

G.先缩点

 #include <queue>
 #include <vector>
 #include <cstring>
 #include <iostream>

 using namespace std;

 const int maxn = ;

 struct node{
     int x, y;
     bool operator < (const node &a) const {
         return y > a.y;
     }
 };

 vector <node> e[maxn], f[maxn];

 priority_queue <node> q;

 int n, N, m, vi[maxn];

 long long ans;

 int cnt, top, in[maxn], dfn[maxn], low[maxn], sta[maxn], vis[maxn], bel[maxn], sum[maxn];

 void Clear() {
     ans = cnt = N = ;
     for(int i = ;i < n;i ++) e[i].clear(), f[i].clear(), vi[i] = vis[i] = ;
     while(!q.empty()) q.pop();
 }

 void tarjan(int x, int t = ) {
     vis[x] = ;
     dfn[x] = low[x] = ++ cnt, sta[++ top] = x, in[x] = ;
     for(int i = ;i < e[x].size();i ++) {
         t = e[x][i].x;
         if(!vis[t]) {
             tarjan(t);
             low[x] = min(low[x], low[t]);
         }
         else if(in[t] && low[t] < low[x]) low[x] = low[t];
     }
     if(low[x] == dfn[x]) {
         for(int i = -;i != x;i = sta[top --], bel[i] = N, in[i] = );
         N ++;
     }
 }

 int main() {
     ios::sync_with_stdio(false);
     int u, v, w;
     node tmp;
     while(cin >> n >> m) {
         Clear();
         for(int i = ;i <= m;i ++)
             cin >> u >> v >> w, e[u].push_back((node){v, w});
         for(int i = ;i < n;i ++)
             if(!vis[i]) tarjan(i);
         memset(sum, 0x3f,sizeof sum);
         for(int i = ;i < n;i ++) {
             for(int j = ;j < e[i].size();j ++)
                 if(bel[i] != bel[e[i][j].x])
                     sum[bel[e[i][j].x]] = min(sum[bel[e[i][j].x]], e[i][j].y);
         }
         for(int i = ;i < N;i ++)
             if(sum[i] != 0x3f3f3f3f)
                 ans += sum[i];
         cout << ans << endl;
     }
     return ;
 }

H.手写一下发现总共只有2 + 4 + 8 + ... + 2^9 = 1022 个数

所以枚举对比一下就好了

 #include <vector>
 #include <iostream>

 using namespace std;

 vector <int> e[];

 int n, sum;

 int main() {
     e[].push_back();
     cin >> n;
     for(int i = , j = ;i <= ;i ++, j *= ) {
         for(int t, k = ;k < e[i - ].size();k ++) {
             t =  * j + e[i - ][k];
             e[i].push_back(t);
             if(t == n) {
                 cout << sum + e[i].size();
                 return ;
             }
         }
         for(int t, k = ;k < e[i - ].size();k ++) {
             t =  * j + e[i - ][k];
             e[i].push_back(t);
             if(t == n) {
                 cout << sum + e[i].size();
                 return ;
             }
         }
         sum += e[i].size();
     }
 }