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

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

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

0

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

5

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

3

題意:n個城鎮，m條道路(無向 s(起點 t(終點 然後求建一條新路，s-t的最短路徑不會改變。
思路:跑兩次spfa s到各個點的最短路 t到各個點的最短路 然後枚舉每次沒有路的點 然後判斷dis1[i]+dis2[j]+1>=dis1[t]&&dis1[j]+dis2[i]+1>=dis1[t]，這樣就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;

}

PS：摸魚怪的博客分享，歡迎感謝各路大牛的指點~

CodeForces - 954D Fight Against Traffic