• Heavy Transportation(POJ


    Background

    Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand

    business. But he needs a clever man who tells him whether there really is a way from the place

    his customer has build his giant steel crane to the place where it is needed on which all

    streets can carry the weight.

    Fortunately he already has a plan of the city with all streets and bridges and all the allowed

    weights.Unfortunately he has no idea how to find the the maximum weight capacity in order

    to tell his customer how heavy the crane may become. But you surely know.

    Problem
    You are given the plan of the city, described by the streets (with weight limits) between the

    crossings, which are numbered from 1 to n. Your task is to find the maximum weight that can

    be transported from crossing 1 (Hugo’s place) to crossing n (the customer’s place). You may

    assume that there is at least one path. All streets can be travelled in both directions.

    Input

    The first line contains the number of scenarios (city plans). For each city the number n of

    street crossings (1 <= n <= 1000) and number m of streets are given on the first line. The

    following m lines contain triples of integers specifying start and end crossing of the street and

    the maximum allowed weight, which is positive and not larger than 1000000. There will be at

    most one street between each pair of crossings.

    Output

    The output for every scenario begins with a line containing “Scenario #i:”, where i is the

    number of the scenario starting at 1. Then print a single line containing the maximum allowed

    weight that Hugo can transport to the customer. Terminate the output for the scenario with a

    blank line.

    Sample Input

    1
    3 3
    1 2 3
    1 3 4
    2 3 5

    Sample Output

    Scenario #1:
    4

    //注意题上最后让输出2个换行
    //#include<map>
    #include<queue>
    #include<stack>
    #include<vector>
    #include<math.h>
    #include<cstdio>
    #include<sstream>
    #include<numeric>//STL数值算法头文件
    #include<stdlib.h>
    #include<string.h>
    #include<iostream>
    #include<algorithm>
    #include<functional>//模板类头文件
    using namespace std;
    
    const int INF=1e9+7;
    const int maxn=1100;
    typedef long long ll;
    
    int t,n,m;
    int vis[maxn];
    int par[maxn];
    int tu[maxn][maxn];
    
    int dijkstra()
    {
        int i,j,k;
        memset(vis,0,sizeof(vis));
        for(i=1; i<=n; i++)
            par[i]=tu[1][i];
        for(i=1; i<=n; i++)
        {
            int minn=-1;
            for(j=1; j<=n; j++)
            {
                if(!vis[j]&&minn<par[j])
                {
                    k=j;
                    minn=par[j];
                }
            }
            vis[k]=1;
            for(j=1; j<=n; j++)
            {
                par[j]=max(par[j],min(tu[k][j],par[k]));
    //            if(!vis[j]&&par[j]<min(tu[k][j],par[k]))
    //            {
    //                par[j]=min(tu[k][j],par[k]);
    //            }
            }
        }
        return par[n];
    }
    
    int main()
    {
        int cot=1;
        scanf("%d",&t);
        while(t--)
        {
            int i,j,a,b,c;
            scanf("%d %d",&n,&m);
            memset(tu,0,sizeof(tu));
            while(m--)
            {
                scanf("%d %d %d",&a,&b,&c);
                tu[a][b]=tu[b][a]=c;
            }
            printf("Scenario #%d:
    ",cot++);
            printf("%d
    
    ",dijkstra());
        }
        return 0;
    }
    
  • 相关阅读:
    深入浅出聊优化:从Draw Calls到GC
    关于Unity中植物树木烘焙后没有影子的解决方法
    Marvelous Designer 服装设计与模拟
    DAZ studio 4.9基础
    在下载SOPC代码的过程中遇到的一些错误
    开发工程师人生之路
    简易信号发生器的设计
    HDU A Simple Math Problem (矩阵快速幂)
    HDU Queuing (递推+矩阵快速幂)
    POJ 3233 Matrix Power Series(矩阵快速幂)
  • 原文地址:https://www.cnblogs.com/nyist-xsk/p/7264838.html
Copyright © 2020-2023  润新知