CodeForces - 954D Fight Against Traffic

Posted 2020-11-10 MengX

Fight Against Traffic

Little town Nsk consists of n junctions connected by m bidirectional roads. Each road connects two distinct junctions and no two roads connect the same pair of junctions. It is possible to get from any junction to any other junction by these roads. The distance between two junctions is equal to the minimum possible number of roads on a path between them.

In order to improve the transportation system, the city council asks mayor to build one new road. The problem is that the mayor has just bought a wonderful new car and he really enjoys a ride from his home, located near junction s to work located near junction t. Thus, he wants to build a new road in such a way that the distance between these two junctions won‘t decrease.

You are assigned a task to compute the number of pairs of junctions that are not connected by the road, such that if the new road between these two junctions is built the distance between s and t won‘t decrease.

Input

The firt line of the input contains integers n, m, s and t (2?≤?n?≤?1000, 1?≤?m?≤?1000, 1?≤?s,?t?≤?n, s?≠?t) — the number of junctions and the number of roads in Nsk, as well as the indices of junctions where mayors home and work are located respectively. The i-th of the following m lines contains two integers u_i and v_i (1?≤?u_i,?v_i?≤?n, u_i?≠?v_i), meaning that this road connects junctions u_i and v_i directly. It is guaranteed that there is a path between any two junctions and no two roads connect the same pair of junctions.

Output

Print one integer — the number of pairs of junctions not connected by a direct road, such that building a road between these two junctions won‘t decrease the distance between junctions s and t.

Examples

Input

5 4 1 5
1 2
2 3
3 4
4 5

Output

0

Input

5 4 3 5
1 2
2 3
3 4
4 5

Output

5

Input

5 6 1 5
1 2
1 3
1 4
4 5
3 5
2 5

Output

3

题意:n个城镇，m条道路(无向 s(起点 t(终点 然后求建一条新路，s-t的最短路径不会改变。
思路:跑两次spfa s到各个点的最短路 t到各个点的最短路 然后枚举每次没有路的点 然后判断dis1[i]+dis2[j]+1>=dis1[t]&&dis1[j]+dis2[i]+1>=dis1[t]，这样就cnt++

#include<bits/stdc++.h>
#define ll long long
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const int maxn= 1000+5;
const int INF = 0x3f3f3f3f;
int dis1[maxn],dis2[maxn];
int d[maxn][maxn],n,m;
bool vis[maxn];
void spfa(int s,int dis[1005])
{
    queue<int> q;
    memset(vis,0,sizeof(vis));
    vis[s]=1;
    dis[s]=0;
    q.push(s);
    while(!q.empty())
    {
        int v=q.front();
        q.pop();
        vis[v]=0;
        for(int i=1; i<=n; i++)
        {
            int w=d[v][i];
            if(w==INF)
                continue;
            if(dis[i]>dis[v]+w)
            {
                dis[i]=dis[v]+w;
                if(!vis[i])
                {
                    q.push(i);
                    vis[i]=1;
                }
            }
        }

    }

}
int main()
{
    int s,t,u,v;
    scanf("%d %d %d %d",&n,&m,&s,&t);
    memset(d,INF,sizeof(d));
    memset(dis1,INF,sizeof(dis1));
    memset(dis2,INF,sizeof(dis2));
    for(int i=1; i<=m; i++)
    {
        scanf("%d %d",&u,&v);
        d[u][v]=d[v][u]=1;
    }

    spfa(s,dis1);
    spfa(t,dis2);
    int cnt=0;
    for(int i=1; i<=n; i++)
        for(int j=1; j<=n&&i!=j; j++)
        {
            if(i==t&&j==s)
                continue;
            if((dis1[i]+dis2[j]+1>=dis1[t])&&(dis1[j]+1+dis2[i]>=dis1[t])&&d[i][j]==INF)
            {
                cnt++;
            }
        }
    cout<<cnt<<endl;

}

View Code

 

 

 

PS：摸鱼怪的博客分享，欢迎感谢各路大牛的指点~