对于每一个城市 i，假定只有第i 个城市与外界交通中断，输出有多少个城市会因此导致到首都的最短路径长度改变。
我们采用邻接表的方式存储图的信息，其中 head[x]表示顶点x 的第一条边的编号，next[i]表示第i 条边的下一条边的编号，point[i]表示第 i 条边的终点，weight[i]表示第 i 条边的长度。（第一空 2 分，其余3 分）
#include <iostream>
#include <cstring>
using namespace std;
#define MAXN 6000
#define MAXM 100000
#define infinity 2147483647
int head[MAXN], next[MAXM], point[MAXM], weight[MAXM];
int queue[MAXN], dist[MAXN], visit[MAXN];
int n, m, x, y, z, total = 0, answer;
void link(int x, int y, int z) {
total++;
next[total] = head[x];
head[x] = total;
point[total] = y;
weight[total] = z;
total++;
next[total] = head[y];
head[y] = total;
point[total] = x;
weight[total] = z;
}
int main( ) {
int i, j, s, t;
cin >> n >> m;
for (i = 1; i <= m; i++) {
cin >> x >> y >> z;
link(x, y, z);
}
for (i = 1; i <= n; i++) dist[i] = infinity;
1;
queue[1] = 1;
visit[1] = 1;
s = 1;
t = 1;
// 使用 SPFA 求出第一个点到其余各点的最短路长度
while (s <= t) {
x = queue[s % MAXN];
j = head[x];
while (j != 0) {
if (2) {
dist[point[j]] = dist[x] + weight[j];
if (visit[point[j]] == 0) {
t++;
queue[t % MAXN] = point[j];
visit[point[j]] = 1;
}
}
j = next[j];
}
3;
s++;
}
for (i = 2; i <= n; i++) {
queue[1] = 1;
memset(visit, 0, sizeof(visit));
visit[1] = 1;
s = 1;
t = 1;
while (s <= t) { // 判断最短路长度是否不变
x = queue[s];
j = head[x];
while (j != 0) {
if (point[j] != i && 4
&& visit[point[j]] == 0) {
5;
t++;
queue[t] = point[j];
}
j = next[j];
}
s++;
}
answer = 0;
for (j = 1; j <= n; j++)
answer += 1 - visit[j];
cout << i << ":" << answer - 1 << endl;
}
return 0;
} 
