POJ #1125 Stockbroker Grapevine Floyd Bfs

Description

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Stockbrokers are known to overreact to rumours. You have been contracted to develop a method of spreading disinformation amongst the stockbrokers to give your employer the tactical edge in the stock market. For maximum effect, you have to spread the rumours in the fastest possible way. 

Unfortunately for you, stockbrokers only trust information coming from their "Trusted sources" This means you have to take into account the structure of their contacts when starting a rumour. It takes a certain amount of time for a specific stockbroker to pass the rumour on to each of his colleagues. Your task will be to write a program that tells you which stockbroker to choose as your starting point for the rumour, as well as the time it will take for the rumour to spread throughout the stockbroker community. This duration is measured as the time needed for the last person to receive the information.

Input

Your program will input data for different sets of stockbrokers. Each set starts with a line with the number of stockbrokers. Following this is a line for each stockbroker which contains the number of people who they have contact with, who these people are, and the time taken for them to pass the message to each person. The format of each stockbroker line is as follows: The line starts with the number of contacts (n), followed by n pairs of integers, one pair for each contact. Each pair lists first a number referring to the contact (e.g. a '1' means person number one in the set), followed by the time in minutes taken to pass a message to that person. There are no special punctuation symbols or spacing rules. 

Each person is numbered 1 through to the number of stockbrokers. The time taken to pass the message on will be between 1 and 10 minutes (inclusive), and the number of contacts will range between 0 and one less than the number of stockbrokers. The number of stockbrokers will range from 1 to 100. The input is terminated by a set of stockbrokers containing 0 (zero) people. 

Output

For each set of data, your program must output a single line containing the person who results in the fastest message transmission, and how long before the last person will receive any given message after you give it to this person, measured in integer minutes. 
It is possible that your program will receive a network of connections that excludes some persons, i.e. some people may be unreachable. If your program detects such a broken network, simply output the message "disjoint". Note that the time taken to pass the message from person A to person B is not necessarily the same as the time taken to pass it from B to A, if such transmission is possible at all.

Sample Input

3
2 2 4 3 5
2 1 2 3 6
2 1 2 2 2
5
3 4 4 2 8 5 3
1 5 8
4 1 6 4 10 2 7 5 2
0
2 2 5 1 5
0

Sample Output

3 2
3 10

思路

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　　依旧是求最短路，由于建图后发现是稠密图且点比较少，所以采用 Floyd，算法的时间复杂度是O(V^3) 。

　　这道题比较不同的地方在于可以同时访问多个邻接点 （相当于群发），这样的话就不是多个结点对最短路径的累加而是求最大值了。

　　最后需要判断是否存在一点可以广度优先遍历全图，如果不存在这样的点则说明有无法联系上的股票经纪人，则输出 ”disjoint“ 。

　　AC代码如下：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
#define MAXN 105
#define MAXM 10000
#define INF 0x3f3f3f3f
int n;
int G[MAXN][MAXN];

void initG () {
    memset(G, INF, sizeof(G));
    int EdgeNum, v, w;
    for (int i = 1; i <= n; ++i) {
        scanf("%d", &EdgeNum);
        for (int j = 1; j <= EdgeNum; ++j) {
            scanf("%d%d", &v, &w);
            if (i != v) G[i][v] = w;
        }
    }
}

//bfs 判断图中是否存在一点，可以广度优先遍历全图
bool bfs (const int& s) {
    bool vis[MAXN];
    memset (vis, false, sizeof(vis));
    queue<int> q; q.push(s);
    vis[s] = true;

    while(!q.empty()) {
        int u = q.front(); q.pop();
        for (int i = 1; i <= n; i++) {
            if (!vis[i] && G[u][i] != INF ) {
                vis[i] = true;
                q.push(i);
            }
        }
    }
    for (int i = 1; i <= n; i++) {
         if (!vis[i]) return false;
    }
    return true;
}

//floyd求所有结点对的最短路
int dp[MAXN][MAXN];
void floyd () {
    memcpy(dp, G, sizeof(G));
    for (int k = 1; k <= n; k++)
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
                if (i != j) dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j]);
}

int main (void) {
    while (scanf("%d", &n) && n) {
        initG();
        bool isCover = false;
        for (int i = 1; i <= n; i++) {
            if (bfs(i)) {
                isCover = true;
                break;
            }
        }
        if (!isCover) {
            printf("disjoint
");
            continue;
        }
        floyd();
        //找到从某一点出发的最短路，可同时访问多个邻接点
        int ans1 = 0, ans2 = INF;
        for (int i = 1; i <= n; i++) {
            int sum = 0;
            for (int j = 1; j <= n; j++) {
                if (i != j) sum = max(sum, dp[i][j]);
            }
            if (ans2 > sum) {
                ans1 = i;
                ans2 = sum;
            }
        }
        printf("%d %d
", ans1, ans2);
    }
    return 0;
}