题目链接 http://codeforces.com/contest/954/problem/D

题目大意

n m s t 分别为点的个数， 边的个数，以及两个特殊的点

要求s与t间的距离在新增一条边下不变

基本思路

用dj算法由s 到 t两个点分别进行一次计算

得出每个点到s与t的最短值

遍历计算每两个没建立联系的边建立联系后，s与t的距离，并与初始时距离比较

若不变则记录（s与t的值必定为新建经由新建这条边的数值或原始值中最小一个）

#include <stdio.h>
#include <algorithm>
#include <queue>
#include <bitset>
#include <memory.h>
#define MS 1010
#define END 0x3f3f3f3f
using namespace std;int a[MS][MS]; //閭绘帴琛?priority_queue <int> q;
void dijsktra(int start, int N, int d[MS]){
int i, j;
bool v[MS];
memset(v, 0, sizeof(v));
d[start] = 0;for(i = 1; i < N; i++){
int x = 0;
for(j = 1; j <= N; j++)
if(!v[j] && (x == 0|| d[j] < d[x])) x = j;
v[x] = 1;/*printf("%d\n", x);*/
for(j = 1; j <= N; j++){
d[j] = min(d[j], d[x]+a[x][j]);
/*printf("%d %d\n", d[x], a[x][j]);*/
}
}
}
int main(){
int N, M, S, T;
int temp_x, temp_y;
while(scanf("%d%d%d%d", &N, &M, &S, &T) == 4){
//init
int i, j;
memset(a, 0x3f, sizeof(a));
for(i = 0; i <= N; i++) a[i][i] = 0;
for(i = 0; i < M; i++){
scanf("%d%d", &temp_x, &temp_y);
a[temp_x][temp_y] = 1;
a[temp_y][temp_x] = 1;
}
/*for(i = 1; i <= N; i++){
for(j = 1; j <= N; j++){
if(a[i][j] > 10)
printf("* ");
else
printf("%d ", a[i][j]);
}
printf("\n");
}*/
int dt[MS], ds[MS];
memset(ds, 0x3f, sizeof(ds));
memset(dt, 0x3f, sizeof(dt));
dijsktra(T, N, dt);
dijsktra(S, N, ds);
int length = ds[T];
/*for(i = 0; i <= N; i++) printf("%d ", dt[i]);
printf("\n");
printf("%d\n", length);*/
int count = 0;
for(i = 1; i <= N; i++){
for(j = i+1; j <= N; j++){
if(a[i][j] == 1) continue;
if(ds[i]+dt[j]+1 >= length && ds[j] + dt[i] +1 >= length){
/*printf("%d %d %d %d\n", i, j, ds[i]+dt[j]+1, ds[j]+dt[i]+1);*/
count++;
}
}
}
printf("%d\n", count);/*for(i = 1; i <= N; i++)
printf("%d ", d[i]);
printf("\n");*/
}
}