题目链接：https://codeforces.com/problemset/problem/954/D
博客园食用链接：https://www.cnblogs.com/lonely-wind-/p/13451829.html
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
n, m, s
n,m,s and
t
(
2
≤
n
≤
1000
,
1
≤
m
≤
1000
,
1
≤
s
,
t
≤
n
,
s
≠
t
)
t (2 ≤ n ≤ 1000, 1 ≤ m ≤ 1000, 1 ≤ s, t ≤ n, s ≠ t)
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
u_ i
ui and
v
i
(
1
≤
u
i
,
v
i
≤
n
,
u
i
≠
v
i
)
v_i (1 ≤ u_ i, v_ i ≤ n, u_i ≠ v i)
vi(1 ≤ ui, vi ≤ n,ui = vi), meaning that this road connects junctions
u
i
u _i
ui and
v
i
v_i
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
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
emmm，题目读的有点迷，不过只要注意到新加一个边和使得 s s s到 t t t的距离不减少（！！神奇的的要求），并且要求新加边的两个点直接之前不能有直达的边，这题就差不多读懂了
题目大意：给你一个图，n个点，m条边，起点和终点 s , t s,t s,t，然后你需要新加一条边在两个点之间，现在问你有多少种加法使得 d i s t ( s , t ) dist(s,t) dist(s,t)不减少。
看题目数据范围，比较小，那么我们可以直接枚举在那两个点之间建边，如果对于点 i , j i,j i,j建边了，若导致了最短路缩减，也就是 d i s ( s , i ) + d i s ( j , t ) < d i s ( s , t ) dis(s,i)+dis(j,t)<dis(s,t) dis(s,i)+dis(j,t)<dis(s,t) 或着 d i s ( s , j ) + d i s ( i , t ) < d i s ( s , t ) dis(s,j)+dis(i,t)<dis(s,t) dis(s,j)+dis(i,t)<dis(s,t)因为没法判断 i , j i,j i,j哪个点离哪个端点最近，所以要判断两次。
如果不减少的话也就是大于等于了，那么我们跑两次最短路就好了，一个对起点跑，一个对终点跑。然后枚举判断一下就好了。对于这种小数据，邻接矩阵他不香吗QAQ。
以下是AC代码：
#include <bits/stdc++.h>
using namespace std;
const int mac=1e3+10;
const int inf=1e9+10;
int mp[mac][mac],diss[mac],n,dist[mac];
int vis[mac];
void dij(int s,int *dis)
{
memset(vis,0,sizeof vis);
for (int i=1; i<=n; i++) dis[i]=mp[s][i];
dis[s]=0; vis[s]=1;
for (int i=1; i<=n; i++){
int minn=inf,k=0;
for (int j=1; j<=n; j++)
if (!vis[j] && dis[j]<minn)
minn=dis[j],k=j;
if (!k) break;
vis[k]=1;
for (int j=1; j<=n; j++)
if (dis[k]+mp[k][j]<dis[j])
dis[j]=dis[k]+mp[k][j];
}
}
int main(int argc, char const *argv[])
{
int m,s,t;
scanf ("%d%d%d%d",&n,&m,&s,&t);
for (int i=1; i<=n; i++)
for (int j=1; j<=n; j++) mp[i][j]=inf;
for (int i=1; i<=m; i++){
int u,v;
scanf ("%d%d",&u,&v);
mp[u][v]=mp[v][u]=1;
}
for (int i=1; i<=n; i++) mp[i][i]=0;
dij(s,diss); dij(t,dist);
int ans=0;
for (int i=1; i<=n; i++){
for (int j=i+1; j<=n; j++){
if (mp[i][j]!=inf) continue;
if (diss[i]+dist[j]+1>=diss[t] && dist[i]+diss[j]+1>=diss[t])
ans++;
}
}
printf("%d\n",ans);
return 0;
}