POJ-1125 Stockbroker Grapevine 最短路

这题要处理的地方的就是一个人可以同时向多个人传递消息，也就是说一条消息的传递时间由最长的那一条路径所决定，因为可以同时进行嘛，所以就求某一点到所有点的最短路，然后再寻找一条最长的路劲，枚举每个顶点作为起点就可以了。

代码如下：

#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<algorithm>

#include<cmath>

#include<queue>

#include<set>

#include<map>

#include<cstring>

#include<vector>

#include<string>

#define inf 0x3f3f3f3f

#define LL long long

using namespace std;



int N, G[105][105], dis[105], vis[105];



bool Dijkstra(int s, int &ret) {

    int pos, Min;

    ret = 0;

    memset(dis, 0x3f, sizeof (dis));

    memset(vis, 0, sizeof (vis));

    dis[s] = 0;

    for (int i = 1; i <= N; ++i) {

        Min = inf;

        for (int j = 1; j <= N; ++j) {

            if (!vis[j] && Min > dis[j]) {

                pos = j, Min = dis[j];

            }

        }

        if (Min == inf) {

            return false;

        } else {

            vis[pos] = 1;

            for (int j = 1; j <= N; ++j) {

                if (!vis[j] && G[pos][j] != inf && dis[j] > dis[pos] + G[pos][j]) {

                    dis[j] = dis[pos] + G[pos][j];    

                }

            }

        }

    }

    for (int i = 1; i <= N; ++i) {

        ret = max(ret, dis[i]);    

    }

    return true;

}



int main()

{

    int a, b, M, flag, Min, num, ret;

    while (scanf("%d", &N), N) {

        flag = 0;

        Min = inf;

        memset(G, 0x3f, sizeof (G));

        for (int i = 1; i <= N; ++i) {

            scanf("%d", &M);

            for (int j = 1; j <= M; ++j) {

                scanf("%d %d", &a, &b);

                G[i][a] = b;

            }

        } 

        for (int i = 1; i <= N; ++i) {

            int t;

            if (Dijkstra(i, t)) {

                flag = 1;

                if (Min > t) {

                    Min = t, num = i;

                }

            }

        }

        if (!flag) puts("disjoint");

        else printf("%d %d\n", num, Min);

    }

    return 0;

}