POJ 1125 Stockbroker Grapevine

题意:有n个人,传播谣言,每个人向其他人传播的时间作为边权,构成一个有向图,问把谣言告诉谁能最快传到所有人,输出这个人和最短时间。

解法:最短路。一个人传到所有人的最短时间即他到所有人最短路的最大值,求所有最大值的最小值即为所求。分别用了floyd和dijsktra写了一下……(一直以为dijsktra就是个广搜而已呢(逃

代码:

floyd

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<math.h>
#include<limits.h>
#include<time.h>
#include<stdlib.h>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define LL long long
using namespace std;

int n;
int dis[105][105];
const int maxn = 0x3f3f3f3f;
void floyd()
{
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= n; j++)
        {
            for(int k = 1; k <= n; k++)
            {
                dis[j][k] = min(dis[j][i] + dis[i][k], dis[j][k]);
            }
        }
    }
}
int main()
{
    while(~scanf("%d", &n) && n)
    {
        memset(dis, maxn, sizeof dis);
        for(int i = 1; i <= n; i++)
        {
            int m;
            scanf("%d", &m);
            for(int j = 0; j < m; j++)
            {
                int a, b;
                scanf("%d%d", &a, &b);
                dis[i][a] = b;
            }
        }
        floyd();
        int ans = INT_MAX, pos = 0;
        for(int i = 1; i <= n; i++)
        {
            int res = -1;
            for(int j = 1; j <= n; j++)
            {
                if(i == j)
                    continue;
                int tmp = dis[i][j];
                if(tmp == maxn)
                {
                    res = -1;
                    break;
                }
                else
                    res = max(res, tmp);
            }
            if(res != -1)
            {
                if(res < ans)
                {
                    ans = res;
                    pos = i;
                }
            }
        }
        if(pos)
        {
            printf("%d %d
", pos, ans);
        }
        else
            puts("disjoint");
    }
    return 0;
}

  

dijsktra

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<math.h>
#include<limits.h>
#include<time.h>
#include<stdlib.h>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define LL long long
using namespace std;

int n;
int stockbroker[105][105];
bool vis[105];
int dis[105][105];
const int maxn = 0x3f3f3f3f;
struct node
{
    int vertex;
    int value;
    node(int vertex, int value) : vertex(vertex), value(value) {}
    node() {}
};
queue <node> q;
void dij()
{
    for(int i = 1; i <= n; i++)
    {
        memset(vis, 0, sizeof vis);
        q.push(node(i, 0));
        while(!q.empty())
        {
            node tmp = q.front();
            q.pop();
            vis[tmp.vertex] = true;
            for(int j = 1; j <= n; j++)
            {
                if(stockbroker[tmp.vertex][j] < 20)
                {
                    dis[i][j] = min(dis[i][j], tmp.value + stockbroker[tmp.vertex][j]);
                    if(!vis[j])
                        q.push(node(j, tmp.value + stockbroker[tmp.vertex][j]));
                }
            }
        }
    }
}
int main()
{
    while(~scanf("%d", &n) && n)
    {
        memset(dis, maxn, sizeof dis);
        for(int i = 0; i < 105; i++)
        {
            for(int j = 0; j < 105; j++)
            {
                stockbroker[i][j] = 20;
                if(i == j)
                    stockbroker[i][j] = 0;
            }
        }
        for(int i = 1; i <= n; i++)
        {
            int m;
            scanf("%d", &m);
            for(int j = 0; j < m; j++)
            {
                int a, b;
                scanf("%d%d", &a, &b);
                stockbroker[i][a] = b;
                dis[i][a] = b;
            }
        }
        dij();
        int ans = INT_MAX, pos = 0;
        for(int i = 1; i <= n; i++)
        {
            int res = -1;
            for(int j = 1; j <= n; j++)
            {
                if(i == j)
                    continue;
                int tmp = dis[i][j];
                if(tmp == maxn)
                {
                    res = -1;
                    break;
                }
                else
                    res = max(res, tmp);
            }
            if(res != -1)
            {
                if(res < ans)
                {
                    ans = res;
                    pos = i;
                }
            }
        }
        if(pos)
        {
            printf("%d %d
", pos, ans);
        }
        else
            puts("disjoint");
    }
    return 0;
}

  

原文地址:https://www.cnblogs.com/Apro/p/4566774.html