ZOJ 3396 Conference Call 求经过特定3点的最小生成树

//ZOJ 3396 Conference Call//题意 求经过特定3点的最小生成树//思路：枚举任何一点作为支撑点 ，特定的3点要相连必须经过共同的一点 ，求这三点到所枚举点的最短路和，取最小值即为答案#include<iostream>#include<stdio.h>#include<string.h>#include<queue>using namespace std;#define INF 50000000#define N 20005#define M 505int n, m, k;int sta[N];int head[M], num;int dis[4][M];bool vis[M], mark[M];struct Edge {    int from;    int to;    int val;    int next;} edge[N];void addedge(int from, int to, int val) {    edge[num].from = from;    edge[num].to = to;    edge[num].val = val;    edge[num].next = head[from];    head[from] = num++;}void init() {    memset(head, -1, sizeof (head));    num = 0;}void spfa(int s, int index) {    memset(vis, 0, sizeof (vis));    for (int i = 1; i <= m; ++i)        dis[index][i] = INF;    queue<int> Q;    Q.push(s);    dis[index][s] = 0;    vis[s] = 1;    while (!Q.empty()) {        int p = Q.front();        Q.pop();        vis[p] = 0;        for (int i = head[p]; i != -1; i = edge[i].next) {            if (dis[index][edge[i].to] > dis[index][p] + edge[i].val) {                dis[index][edge[i].to] = dis[index][p] + edge[i].val;                if (!vis[edge[i].to]) {                    Q.push(edge[i].to);                    vis[edge[i].to] = 1;                }            }        }    }}int main() {    int i, j, ca = 1;    int a, b, c;    while (scanf("%d %d %d", &n, &m, &k) != EOF) {        init();        for (i = 1; i <= n; ++i)            scanf("%d", &sta[i]);        for (i = 1; i <= k; ++i) {            scanf("%d %d %d", &a, &b, &c);                   addedge(a, b, c);            addedge(b, a, c);        }        int bx;        int qua;        scanf("%d", &qua);        printf("Case #%d\n", ca++);        for (i = 1; i <= qua; ++i) {            scanf("%d %d %d", &a, &b, &c);            spfa(sta[a], 1);            spfa(sta[b], 2);            spfa(sta[c], 3);            int ans = INF;            for (j = 1; j <= m; ++j)                if (dis[1][j] + dis[2][j] + dis[3][j] < ans){                    ans = dis[1][j] + dis[2][j] + dis[3][j];                    bx = j;                }            printf("Line %d: ", i);           // printf("bx :%d %d %d %d\n",bx,dis[1][bx],dis[2][bx],dis[3][bx]);            if (ans == INF)                printf("Impossible to connect!\n");            else                printf("The minimum cost for this line is %d.\n", ans);        }    }    return 0;}