题目链接:
http://codeforces.com/contest/544/problem/D
题意:
有n个城镇，m条边权为1的双向边
让你破坏最多的道路，使得从s1到t1，从s2到t2的距离分别不超过l1和l2
题解:
跑一发最短路，然后最后留下的图肯定是出了s1-t1,s2-t2这两条路之外，其他路都被删除了
那么我们枚举重叠的道路就好了【可能反向】
代码:
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 #define MS(a) memset(a,0,sizeof(a)) 5 #define MP make_pair 6 #define PB push_back 7 const int INF = 0x3f3f3f3f; 8 const ll INFLL = 0x3f3f3f3f3f3f3f3fLL; 9 inline ll read(){ 10 ll x=0,f=1;char ch=getchar(); 11 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 12 while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 13 return x*f; 14 } 15 ////////////////////////////////////////////////////////////////////////// 16 const int maxn = 3e3+10; 17 18 int inq[maxn],d[maxn][maxn]; 19 vector<int> E[maxn]; 20 21 int main(){ 22 int n=read(),m=read(); 23 for(int i=0; i<m; i++){ 24 int u=read(),v=read(); 25 E[u].push_back(v); 26 E[v].push_back(u); 27 } 28 29 int s1,t1,l1,s2,t2,l2; 30 scanf("%d%d%d%d%d%d",&s1,&t1,&l1,&s2,&t2,&l2); 31 32 for(int i=1; i<=n; i++){ 33 MS(inq); 34 for(int j=0; j<=n; j++) d[i][j] = INF; 35 queue<int> q; 36 q.push(i),inq[i]=1,d[i][i]=0; 37 while(!q.empty()){ 38 int now = q.front(); 39 q.pop(),inq[now] = 0; 40 for(int k=0; k<(int)E[now].size(); k++){ 41 int v = E[now][k]; 42 if(d[i][v] > d[i][now]+1){ 43 d[i][v] = d[i][now]+1; 44 if(inq[v]) continue; 45 inq[v] = 1; 46 q.push(v); 47 } 48 } 49 } 50 } 51 52 if(d[s1][t1]>l1 || d[s2][t2]>l2){ 53 cout << -1 << endl; 54 return 0; 55 } 56 57 int ans = d[s1][t1]+d[s2][t2]; 58 for(int i=1; i<=n; i++) 59 for(int j=1; j<=n; j++){ 60 if(d[s1][i]+d[i][j]+d[j][t1]<=l1 && d[s2][i]+d[i][j]+d[j][t2]<=l2) 61 ans = min(ans,d[s1][i]+d[i][j]+d[j][t1]+d[s2][i]+d[j][t2]); 62 if(d[s1][i]+d[i][j]+d[j][t1]<=l1 && d[t2][i]+d[i][j]+d[j][s2]<=l2) // 反向的时候 63 ans = min(ans,d[s1][i]+d[i][j]+d[j][t1]+d[t2][i]+d[j][s2]); 64 } 65 66 cout << m-ans << endl; 67 68 return 0; 69 } 70 /* 71 10 11 72 1 3 73 2 3 74 3 4 75 4 5 76 4 6 77 3 7 78 3 8 79 4 9 80 4 10 81 7 9 82 8 10 83 1 5 3 84 6 2 3 85 */