浙江财经大学
信息管理与工程学院

HDU 3665 Seaside(Floyd)

本文由 Ocrosoft 于 2016-08-07 21:39:37 发表

Seaside

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1575    Accepted Submission(s): 1148

Problem Description
XiaoY is living in a big city, there are N towns in it and some towns near the sea. All these towns are numbered from 0 to N-1 and XiaoY lives in the town numbered ’0’. There are some directed roads connecting them. It is guaranteed that you can reach any town from the town numbered ’0’, but not all towns connect to each other by roads directly, and there is no ring in this city. One day, XiaoY want to go to the seaside, he asks you to help him find out the shortest way.
 

Input
There are several test cases. In each cases the first line contains an integer N (0<=N<=10), indicating the number of the towns. Then followed N blocks of data, in block-i there are two integers, Mi (0<=Mi<=N-1) and Pi, then Mi lines followed. Mi means there are Mi roads beginning with the i-th town. Pi indicates whether the i-th town is near to the sea, Pi=0 means No, Pi=1 means Yes. In next Mi lines, each line contains two integers SMi and LMi, which means that the distance between the i-th town and the SMi town is LMi.
 

Output
Each case takes one line, print the shortest length that XiaoY reach seaside.
 

Sample Input
	
5 1 0 1 1 2 0 2 3 3 1 1 1 4 100 0 1 0 1
 

Sample Output
	
2
 

Solution

题意:有一些城市,要从城市0,走到任意一个靠海的城市。

输入先是一个N,表示城市数量,接下来N块,每块开始两个整数表示这个城市的道路数量M、这个城市是否靠海。接下来M行表示这个城市能到达的城市的编号、距离。

题解:增加一个标记,最后找出城市0到所有有标记的(海边的)城市的最短路径即可。

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <stack>
#include <queue>
#include <ctime>
#include <string>
#include <cstdio>
#include <vector>
#include <cctype>
#include <climits>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define ms(a) memset(a,0,sizeof(a))
#define INF INT_MAX
#define MAXN 150+10
using namespace std;
int edge[MAXN][MAXN];
bool seaside[MAXN];
int main()
{
    int n,m;
    while(cin>>n)
    {
        ms(seaside);
        for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)edge[i][j]=INF;
        for(int i=0;i<n;i++)
        {
            cin>>m>>seaside[i];
            for(int j=0;j<m;j++)
            {
                int w,v;
                cin>>v>>w;
                if(edge[i][v]>w)edge[i][v]=edge[v][i]=w;
            }
        }
        //Floyd
        for(int k=0;k<n;k++)
            for(int i=0;i<n;i++)
                for(int j=0;j<n;j++)
                    if(edge[i][k]<INF&&edge[k][j]<INF)
                        edge[i][j]=min(edge[i][j],edge[i][k]+edge[k][j]);
        int minn=INF;
        for(int i=0;i<n;i++)
        {
            if(seaside[i]&&!i)minn=0;
            else if(seaside[i])minn=min(minn,edge[0][i]);
        }
        printf("%d\n",minn);
    }
    return 0;
}

 

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