[CCF2015.09]题解

201509-1 数列分段

水，记下前一个数，看看跟当前是否一样，不一样就ans+1

 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <fstream>
 #include <cassert>
 #include <cstdio>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>

 using namespace std;

 int n, a;

 int main() {
     // freopen("in", "r", stdin);
     while(~scanf("%d", &n)) {
         scanf("%d", &a);
         int cur = a, cnt = ;
         for(int i = ; i < n; i++) {
             scanf("%d", &a);
             if(cur != a) {
                 cnt++;
                 cur = a;
             }
         }
         printf("%d\n", cnt);
     }
     return ;
 }

1

201509-2 日期计算

打表，注意细节就行

 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <fstream>
 #include <cassert>
 #include <cstdio>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>

 using namespace std;

 const int com[] = {,,,,,,,,,,,, };
 const int lep[] = {,,,,,,,,,,,, };
 int y, c, m, d;

 int main() {
     // freopen("in", "r", stdin);
     while(~scanf("%d %d", &y, &c)) {
         d = ;
         if((y %  ==  && y %  != ) || (y %  == )) {
             for(int i = ; i <= ; i++) {
                 if(lep[i] < c) {
                     m = i;
                 }
                 else break;
             }
             d = c - lep[m];
             m++;
             if(d != ) printf("%d\n%d\n", m, d);
             else printf("%d\n%d\n", m, lep[m]-lep[m-]);
         }
         else {
             for(int i = ; i <= ; i++) {
                 if(com[i] < c) {
                     m = i;
                 }
                 else break;
             }
             d = c - com[m];
             m++;
             if(d != ) printf("%d\n%d\n", m, d);
             else printf("%d\n%d\n", m, com[m]-com[m-]);
         }
     }
     return ;
 }

2

201509-3 模版生成系统

字符串大模拟，无耻地大量使用了STL，甚至出现了vector<pair<vector<int>,vector<int> > >这样的结构。思路就是首先定位此行输入的字符串的需要替换的变量起止位置，再记录当前字符串。接下来读取模式串的键值，放到set中。接下来替换。要注意使用string中的replace时候会让字符串长度发生变化，这时候只要从末尾开始匹配，最后再输出就行了。

 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <fstream>
 #include <cassert>
 #include <cstdio>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>

 using namespace std;
 typedef vector<int>::iterator it;
 typedef vector<vector<int> >::iterator iit;
 typedef pair<vector<int>,vector<int> > pvv;
 const int maxn = ;

 int n, m;
 vector<int> start, end;
 vector<pvv> sig;
 vector<string> str;
 map<string, string> var;
 char tmp[maxn];

 int main() {
     // freopen("in", "r", stdin);
     while(~scanf("%d %d", &m, &n)) {
         sig.clear();
         var.clear();
         getchar();
         for(int i = ; i < m; i++) {
             start.clear();
             end.clear();
             gets(tmp);
             int len = strlen(tmp);
             for(int j = ; j < len; j++) {
                 if(tmp[j] == ' ' && tmp[j-] == '{' && tmp[j-] == '{') start.push_back(j+);
                 if(tmp[j] == ' ' && tmp[j+] == '}' && tmp[j+] == '}') end.push_back(j-);
             }
             sig.push_back(pvv(start, end));
             str.push_back(string(tmp));
         }
         for(int i = ; i < n; i++) {
             gets(tmp);
             string raw(tmp);
             int j;
             for(j = ; j < raw.length(); j++) {
                 if(raw[j] == ' ') break;
             }
             var[raw.substr(, j)] = raw.substr(j+, raw.length()-j-);
         }
         for(int i = ; i < sig.size(); i++) {
             if(sig[i].first.empty()) {
                 printf("%s\n", str[i].c_str());
                 continue;
             }
             for(int j = sig[i].first.size() - ; j >= ; j--) {
                 str[i].replace(
                     sig[i].first[j]-, sig[i].second[j]-sig[i].first[j]+,
                     var[str[i].substr(sig[i].first[j], sig[i].second[j]-sig[i].first[j]+)]);
                 // cout << sig[i].first[j] << " " << sig[i].second[j] << endl;
                 // cout << str[i].substr(sig[i].first[j], sig[i].second[j]-sig[i].first[j] + 1) << endl;
             }
             printf("%s\n", str[i].c_str());
         }
     }
     return ;
 }

3

201509-4 高速公路

求多少个连通对。先tarjan跑出所有连通分量，然后枚举任意两个不相等的点，看看是否属于同一个连通分量里。

 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <fstream>
 #include <cassert>
 #include <cstdio>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>

 using namespace std;

 const int maxn = ;
 const int maxm = ;
 typedef struct Edge {
     int u;
     int v;
     int next;
     Edge() { next = -; }
 }Edge;

 int head[maxn], ecnt;
 Edge edge[maxm];
 int n, m;

 int bcnt, dindex;
 int dfn[maxn], low[maxn];
 int stk[maxn], top;
 int belong[maxn];
 bool instk[maxn];

 void init() {
     memset(edge, , sizeof(edge));
     memset(head, -, sizeof(head));
     memset(instk, , sizeof(instk));
     memset(dfn, , sizeof(dfn));
     memset(low, , sizeof(low));
     memset(belong, , sizeof(belong));
     ecnt = top = bcnt = dindex = ;
 }

 void adde(int uu, int vv) {
     edge[ecnt].u = uu;
     edge[ecnt].v = vv;
     edge[ecnt].next = head[uu];
     head[uu] = ecnt++;
 }

 void tarjan(int u) {
     int v = u;
     dfn[u] = low[u] = ++dindex;
     stk[++top] = u;
     instk[u] = ;
     for(int i = head[u]; ~i; i=edge[i].next) {
         v = edge[i].v;
         if(!dfn[v]) {
             tarjan(v);
             low[u] = min(low[u], low[v]);
         }
         else if(instk[v] && dfn[v] < low[u]) {
             low[u] = dfn[v];
         }
     }
     if(dfn[u] == low[u]) {
         bcnt++;
         do {
             v = stk[top--];
             instk[v] = ;
             belong[v] = bcnt;
         } while(v != u);
     }
 }

 int main() {
     // freopen("in", "r", stdin);
     int uu, vv;
     while(~scanf("%d %d", &n, &m)) {
         init();
         for(int i = ; i < m; i++) {
             scanf("%d %d", &uu, &vv);
             adde(uu, vv);
         }
         for(uu = ; uu <= n; uu++) {
             if(!dfn[uu]) {
                 tarjan(uu);
             }
         }
         int ans = ;
         for(int i = ; i <= n; i++) {
             for(int j = i + ; j <= n; j++) {
                 if(belong[i] == belong[j]) {
                     ans++;
                 }
             }
         }
         printf("%d\n", ans);
     }
     return ;
 }

4