codeforces 954D dijkstra變形

Codeforces 954D

Fight Against Traffic

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

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 ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi), meaning that this road connects junctions ui and vi 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
intput
5 4 1 5
1 2
2 3
3 4
4 5
output
0
intput
5 4 3 5
1 2
2 3
3 4
4 5
output
5
intput
5 6 1 5
1 2
1 3
1 4
4 5
3 5
2 5
output
3
           

解題思路

一道典型的dijsktra算法題，先用兩次dijsktra分别算出原點s和終點t到其他各點的距離，求出distance(s,t)，及s與t的距離。再判斷每一對未直接相鄰的點，及 if(d1[i]+d2[j]+1>=num && d1[j]+d2[i]+1>=num)。滿足條件，則數量cnt加一。最後輸出總數cnt；

代碼如下

#include <iostream>
#include <cstdio>
#include <cmath>
#include <set>
#include <string>
#include <queue>
#include <map>
#include <vector>
#include <cstring>
#include <algorithm>
#include <sstream>
#define INF 1000000
#define MAXN 2e5+9
const long long mod=1e6+3;
using namespace std;
int g[1005][1005];
//int via[1005][1005];
int via[1005];
int d1[1005],d2[1005];
int s,t,n,m;
void disk(int* d,int k)
{
    memset(via,0,sizeof(via));
    int i,j;
    int x,y;
    d[k]=0;
    for(i=0;i<n;i++)
    {
        int mmax=INF;
        for(j=1;j<=n;j++)
        {
            if(via[j]) continue;
            if(d[j]<mmax)
            {
                mmax=d[j];
                x=j;
            }
        }
        via[x]=1;
        for(j=1;j<=n;j++)
        {
            if(j==x) continue;
            if(!g[x][j]) continue;
            if(d[x]+1<d[j])
            {
                d[j]=d[x]+1;
            }
        }
    }
}
int main()
{
    scanf("%d%d%d%d",&n,&m,&s,&t);
    int i,j;
    int x,y;
    memset(g,0,sizeof(g));
    memset(d2,INF,sizeof(d2));
    memset(d1,INF,sizeof(d1));
    for(i=0;i<m;i++)
    {
        scanf("%d%d",&x,&y);
        g[x][y]=g[y][x]=1;
    }
    //cout << 1 << endl;
    disk(d1,s);
    disk(d2,t);
    /*for(i=1;i<=n;i++)
        cout << d1[i]<<' ';
    cout << endl;
    for(i=1;i<=n;i++)
        cout << d2[i]<<' ';
    cout << endl;*/
    int num=d1[t];
    int cnt=0;
    //cout << num << endl;
    for(i=1;i<n;i++)
    {
        for(j=i+1;j<=n;j++)
        {
            if((i==s && j==t) || (i==t && j==s)) continue;
            if(g[i][j]) continue;
            if(d1[i]+d2[j]+1>=num && d1[j]+d2[i]+1>=num)
            {
                cnt++;
                //cout <<d1[i] <<' ' <<d2[j] << endl;
                //cout <<i <<' ' <<j << endl;
            }
        }
    }
    printf("%d\n",cnt);
}