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  • 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;
    }
    

      

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