codeforces 544 D Destroying Roads 【最短路】

题意：给出n个点，m条边权为1的无向边，破坏最多的道路，使得从s1到t1,s2到t2的距离不超过d1,d2

因为最后s1,t1是连通的，且要破坏掉最多的道路，那么就是求s1到t1之间的最短路

用bfs求出任意两个顶点之间的距离， 如果d[s1][t1]>d1||d[s2][t2]>d2，那么不可能

然后就枚举重叠的区间（就像题解里面说的"H"形一样）

如果枚举区间是1--n,1--i的话，需要交换四次

如果枚举区间是1--n,1--n的话，则只需要交换一次就可以了

看的这一篇题解：http://www.cnblogs.com/qscqesze/p/4487498.html

交换一次的

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include <cmath>
 #include<stack>
 #include<vector>
 #include<map>
 #include<set>
 #include<queue>
 #include<algorithm>
 using namespace std;

 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)

 typedef long long LL;
 const int INF = (<<)-;
 const int mod=;
 const int maxn=;

 int n,m;
 vector<int> e[maxn];
 int vis[maxn];
 int d[maxn][maxn];

 int main(){
     scanf("%d %d",&n,&m);
     for(int i=;i<=m;i++){
         int u,v;
         scanf("%d %d",&u,&v);
         e[u].push_back(v);
         e[v].push_back(u);
     }

     int s1,t1,d1,s2,t2,d2;
     scanf("%d %d %d %d %d %d",&s1,&t1,&d1,&s2,&t2,&d2);

     for(int i=;i<=n;i++){
         queue<int> q;
         memset(vis,,sizeof(vis));
         vis[i]=;
         q.push(i);

         while(!q.empty()){
             int u=q.front();q.pop();

             for(int j=;j<e[u].size();j++){
                 int v=e[u][j];
                 if(vis[v]) continue;
                 vis[v]=;
                 d[i][v]=d[i][u]+;
                 q.push(v);
             }
         }
     }

     if(d[s1][t1]>d1||d[s2][t2]>d2) {
         puts("-1");
         return ;
     }

     int ans=d[s1][t1]+d[s2][t2];
     for(int i=;i<=n;i++){
         for(int j=;j<=n;j++){
             if(d[s1][i]+d[i][j]+d[j][t1]<=d1&&d[s2][i]+d[i][j]+d[j][t2]<=d2)
                 ans=min(ans,d[s1][i]+d[i][j]+d[j][t1]+d[s2][i]+d[j][t2]);

             if(d[s1][i]+d[i][j]+d[j][t1]<=d1&&d[t2][i]+d[i][j]+d[j][s2]<=d2)
                 ans=min(ans,d[s1][i]+d[i][j]+d[j][t1]+d[t2][i]+d[j][s2]);
         }
     }
     printf("%d\n",m-ans);
     return ;
 }

交换四次的

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include <cmath>
 #include<stack>
 #include<vector>
 #include<map>
 #include<set>
 #include<queue>
 #include<algorithm>
 using namespace std;

 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)

 typedef long long LL;
 const int INF = (<<)-;
 const int mod=;
 const int maxn=;

 int n,m;
 vector<int> e[maxn];
 int vis[maxn];
 int d[maxn][maxn];

 int main(){
     scanf("%d %d",&n,&m);
     for(int i=;i<=m;i++){
         int u,v;
         scanf("%d %d",&u,&v);
         e[u].push_back(v);
         e[v].push_back(u);
     }

     int s1,t1,d1,s2,t2,d2;
     scanf("%d %d %d %d %d %d",&s1,&t1,&d1,&s2,&t2,&d2);

     for(int i=;i<=n;i++){
         queue<int> q;
         memset(vis,,sizeof(vis));
         vis[i]=;
         q.push(i);

         while(!q.empty()){
             int u=q.front();q.pop();

             for(int j=;j<e[u].size();j++){
                 int v=e[u][j];
                 if(vis[v]) continue;
                 vis[v]=;
                 d[i][v]=d[i][u]+;
                 q.push(v);
             }
         }
     }

     if(d[s1][t1]>d1||d[s2][t2]>d2) {
         puts("-1");
         return ;
     }

     int ans=d[s1][t1]+d[s2][t2];
     for(int i=;i<=n;i++){
         for(int j=;j<=i;j++){
             if(d[s1][i]+d[i][j]+d[j][t1]<=d1&&d[s2][i]+d[i][j]+d[j][t2]<=d2)
                 ans=min(ans,d[s1][i]+d[i][j]+d[j][t1]+d[s2][i]+d[j][t2]);

             if(d[s1][i]+d[i][j]+d[j][t1]<=d1&&d[t2][i]+d[i][j]+d[j][s2]<=d2)
                 ans=min(ans,d[s1][i]+d[i][j]+d[j][t1]+d[t2][i]+d[j][s2]);    

             if(d[t1][i]+d[i][j]+d[j][s1]<=d1&&d[s2][i]+d[i][j]+d[j][t2]<=d2)
                 ans=min(ans,d[t1][i]+d[i][j]+d[j][s1]+d[s2][i]+d[j][t2]);

             if(d[t1][i]+d[i][j]+d[j][s1]<=d1&&d[t2][i]+d[i][j]+d[j][s2]<=d2)
                 ans=min(ans,d[t1][i]+d[i][j]+d[j][s1]+d[t2][i]+d[j][s2]);
         }
     }
     printf("%d\n",m-ans);
     return ;
 }

加油---g00000000----