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:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MS(a) memset(a,0,sizeof(a))
#define MP make_pair
#define PB push_back
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
inline ll read(){
ll x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
//////////////////////////////////////////////////////////////////////////
const int maxn = 3e3+10;
int inq[maxn],d[maxn][maxn];
vector<int> E[maxn];
int main(){
int n=read(),m=read();
for(int i=0; i<m; i++){
int u=read(),v=read();
E[u].push_back(v);
E[v].push_back(u);
}
int s1,t1,l1,s2,t2,l2;
scanf("%d%d%d%d%d%d",&s1,&t1,&l1,&s2,&t2,&l2);
for(int i=1; i<=n; i++){
MS(inq);
for(int j=0; j<=n; j++) d[i][j] = INF;
queue<int> q;
q.push(i),inq[i]=1,d[i][i]=0;
while(!q.empty()){
int now = q.front();
q.pop(),inq[now] = 0;
for(int k=0; k<(int)E[now].size(); k++){
int v = E[now][k];
if(d[i][v] > d[i][now]+1){
d[i][v] = d[i][now]+1;
if(inq[v]) continue;
inq[v] = 1;
q.push(v);
}
}
}
}
if(d[s1][t1]>l1 || d[s2][t2]>l2){
cout << -1 << endl;
return 0;
}
int ans = d[s1][t1]+d[s2][t2];
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++){
if(d[s1][i]+d[i][j]+d[j][t1]<=l1 && d[s2][i]+d[i][j]+d[j][t2]<=l2)
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]<=l1 && d[t2][i]+d[i][j]+d[j][s2]<=l2) // 反向的时候
ans = min(ans,d[s1][i]+d[i][j]+d[j][t1]+d[t2][i]+d[j][s2]);
}
cout << m-ans << endl;
return 0;
}
/*
10 11
1 3
2 3
3 4
4 5
4 6
3 7
3 8
4 9
4 10
7 9
8 10
1 5 3
6 2 3
*/
