训练赛之 最短路

转自：

http://blog.youkuaiyun.com/chenzhenyu123456/article/details/49758663

题意：给定n个点m条边的无向图(边权全为1)，让你去掉最多的边使得d(s1, t1) <= l1 && d(s2, t2) <= l2，若不能满足输出-1，反之输出可以去掉的最多边数。

思路：SPFA预处理所有点之间的距离。求出在满足d(s1, t1) <= l1 && d(s2, t2) <= l2的前提下，路径需要的最少边数ans，答案就是m - ans。

方法是：用dist[i][j]存储最短路。枚举d(s1, t1) 和 d(s2, t2)这两条路径上可能重合的路径d(i, j)

(1)D1=dist[s1][i] + dist[i][j] + dist[j][t1] <= l1 && D2=dist[s2][i] + dist[i][j] + dist[j][t2] <= l2

(2)D1=dist[s1][i] + dist[i][j] + dist[j][t1] <= l1 && D2=dist[s2][j] + dist[j][i] + dist[i][t2] <= l2

(3)D1=dist[s1][j] + dist[j][i] + dist[i][t1] <= l1 && D2=dist[s2][i] + dist[i][j] + dist[j][t2] <= l2

(4)D1=dist[s1][j] + dist[j][i] + dist[i][t1] <= l1 && D2=dist[s2][j] + dist[j][i] + dist[i][t2] <= l2

更新答案为ans = min(ans,  D1 + D2 - dist[i][j])。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<string>
#include<cstring>
#include<iomanip>
#include<iostream>
#include<stack>
#include<cmath>
#include<map>
#include<vector>
#define inf 0x3f3f3f3f
#define INF 1000000000
#define bug1 cout<<"bug1"<<endl;
#define bug2 cout<<"bug2"<<endl;
#define bug3 cout<<"bug3"<<endl;
using namespace std;
typedef long long ll;
const int maxn = 3000 + 5;
int n,m;
int dist[maxn][maxn];int tol;
int head[maxn];
int vis[maxn];
struct Edge{
    int v,nxt;
}edge[maxn*2];
void addedge(int u,int v){
    edge[tol].v=v;edge[tol].nxt=head[u];
    head[u]=tol++;
}
void spfa(int s,int *d){
    queue<int>q;
    memset(vis,0,sizeof(vis));
    q.push(s);d[s]=0;
    while(!q.empty()){
        int u=q.front();q.pop();
        vis[u]=0;
        for(int i=head[u];i!=-1;i=edge[i].nxt){
            int v=edge[i].v;
            if(d[v]>d[u]+1){
                d[v]=d[u]+1;
                if(!vis[v]){
                    vis[v]=1;
                    q.push(v);
                }
            }
        }
    }
}
void init(){
    memset(head,-1,sizeof(head));
    memset(dist,inf,sizeof(dist));
    tol=0;
}
int main()
{
    while(~scanf("%d%d",&n,&m)) {
            init();
        for(int i = 1; i <= m; i++) {
            int u,v;scanf("%d%d",&u,&v);
            addedge(u,v);
            addedge(v,u);
        }
        for(int i = 1; i <= n; i++)spfa(i,dist[i]);
        int s1,t1,l1,s2,t2,l2;
        scanf("%d%d%d%d%d%d",&s1,&t1,&l1,&s2,&t2,&l2);
        if(dist[s1][t1]>l1||dist[s2][t2]>l2){
            printf("-1\n");
            continue;
        }
        int res = dist[s1][t1]+dist[s2][t2];
        for(int i = 1; i <= n; i++) {
            for(int j = i; j <= n; j++) {
                int w1 = min(dist[s1][i]+dist[i][j]+dist[j][t1], dist[s1][j]+dist[j][i]+dist[i][t1]);
                int w2 = min(dist[s2][i]+dist[i][j]+dist[j][t2], dist[s2][j]+dist[j][i]+dist[i][t2]);
                if(w1>l1||w2>l2)continue;
                res = min(res,w1+w2-dist[i][j]);
            }
        }
        printf("%d\n", m - res);
    }
    return 0;
}