51Nod1444 破坏道路
也是非常套路的一道题
首先考虑,如果只有一组限制,那么答案就是m-l1
现在考虑加了一组限制的情况下有什么影响
显然,被两段路径重复覆盖的那一部分会多减掉一次
那么我们可以枚举这段路径,让后用dp来计算答案,注意因为是无权图可以bfs O(n^2)求出两点之间最短路
#pragma GCC opitmize("O3")
#pragma G++ opitmize("O3")
#include<queue>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define N 3010
using namespace std;
struct edge{ int v,nt; } G[6010];
int d[N][N],n,m,h[N],cnt=0; queue<int> q;
inline void gmin(int& x,int y){ x>y?x=y:0; }
int main(){
scanf("%d%d",&n,&m);
for(int x,y,i=1;i<=m;++i){
scanf("%d%d",&x,&y);
G[++cnt]=(edge){y,h[x]}; h[x]=cnt;
G[++cnt]=(edge){x,h[y]}; h[y]=cnt;
}
for(int i=1;i<=n;++i){
bool vis[N]={0};
d[i][i]=0; q.push(i); vis[i]=1;
for(int x;!q.empty();q.pop()){
x=q.front();
for(int v,j=h[x];j;j=G[j].nt)
if(!vis[v=G[j].v]){ d[i][v]=d[i][x]+1; vis[v]=1; q.push(v); }
}
}
int s1,t1,s2,t2,l1,l2,ans;
scanf("%d%d%d%d%d%d",&s1,&t1,&l1,&s2,&t2,&l2); ans=d[s1][t1]+d[s2][t2];
if(d[s1][t1]>l1 || d[s2][t2]>l2) return 0&puts("-1");
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) gmin(ans,d[s1][i]+d[j][t1]+d[i][j]+d[s2][i]+d[j][t2]);
if(d[t1][i]+d[i][j]+d[j][s1]<=l1 && d[s2][i]+d[i][j]+d[j][t2]<=l2) gmin(ans,d[t1][i]+d[j][s1]+d[i][j]+d[s2][i]+d[j][t2]);
if(d[t1][i]+d[i][j]+d[j][s1]<=l1 && d[t2][i]+d[i][j]+d[j][s2]<=l2) gmin(ans,d[t1][i]+d[j][s1]+d[i][j]+d[t2][i]+d[j][s2]);
if(d[s1][i]+d[i][j]+d[j][t1]<=l1 && d[t2][i]+d[i][j]+d[j][s2]<=l2) gmin(ans,d[s1][i]+d[j][t1]+d[i][j]+d[t2][i]+d[j][s2]);
}
printf("%d\n",m-ans);
}