zoukankan      html  css  js  c++  java
  • HDOJ 5418 Victor and World 状压DP


    水状压DP

    Victor and World

    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)
    Total Submission(s): 407    Accepted Submission(s): 164


    Problem Description
    After trying hard for many years, Victor has finally received a pilot license. To have a celebration, he intends to buy himself an airplane and fly around the world. There are n countries on the earth, which are numbered from 1 to n. They are connected by m undirected flights, detailedly the i-th flight connects the ui-th and the vi-th country, and it will cost Victor's airplane wi L fuel if Victor flies through it. And it is possible for him to fly to every country from the first country.

    Victor now is at the country whose number is 1, he wants to know the minimal amount of fuel for him to visit every country at least once and finally return to the first country.
     

    Input
    The first line of the input contains an integer T, denoting the number of test cases.
    In every test case, there are two integers n and m in the first line, denoting the number of the countries and the number of the flights.

    Then there are m lines, each line contains three integers uivi and wi, describing a flight.

    1T20.

    1n16.

    1m100000.

    1wi100.

    1ui,vin.
     

    Output
    Your program should print T lines : the i-th of these should contain a single integer, denoting the minimal amount of fuel for Victor to finish the travel.
     

    Sample Input
    1 3 2 1 2 2 1 3 3
     

    Sample Output
    10
     

    Source
     



    /* ***********************************************
    Author        :CKboss
    Created Time  :2015年08月23日 星期日 10时27分21秒
    File Name     :HDOJ5418.cpp
    ************************************************ */
    
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <string>
    #include <cmath>
    #include <cstdlib>
    #include <vector>
    #include <queue>
    #include <set>
    #include <map>
    #include <cassert>
    
    using namespace std;
    
    typedef long long int LL;
    
    const int INF=0x3f3f3f3f;
    
    int n,m;
    int G[18][18];
    int dp[18][1<<18];
    
    int main()
    {
    	//freopen("in.txt","r",stdin);
    	//freopen("out.txt","w",stdout);
    
    	int T_T;
    	scanf("%d",&T_T);
    	while(T_T--)
    	{
    		scanf("%d%d",&n,&m);
    		memset(G,63,sizeof(G));
    		for(int i=0;i<n;i++) G[i][i]=0;
    		for(int i=0,a,b,c;i<m;i++)
    		{
    			scanf("%d%d%d",&a,&b,&c);
    			a--; b--;
    			G[a][b]=min(G[a][b],c);
    			G[b][a]=min(G[b][a],c);
    		}
    
    		for(int k=0;k<n;k++)
    			for(int i=0;i<n;i++)
    				for(int j=0;j<n;j++)
    					G[i][j]=min(G[i][j],G[i][k]+G[k][j]);
    		memset(dp,63,sizeof(dp));
    
    		dp[0][1]=0;
    		for(int i=1;i<(1<<n);i++) /// status
    		{
    			for(int j=0;j<n;j++) /// from point
    			{
    				if(((1<<j)&i)!=0)
    				{
    					for(int k=0;k<n;k++) /// to point
    					{
    						if(((1<<k)&i)==0)
    						{
    							dp[k][((1<<k)|i)]=min(dp[k][((1<<k)|i)],dp[j][i]+G[j][k]);
    						}
    					}
    				}
    			}
    		}
    		int ans=INF;
    		for(int i=0;i<n;i++)
    		{
    			ans=min(ans,dp[i][(1<<n)-1]+G[i][0]);
    		}
    		if(n==1) ans=0;
    		printf("%d
    ",ans);
    	}
        return 0;
    }
    


  • 相关阅读:
    LibreOJ 6282 数列分块入门 6(在线插入在线查询)
    LibreOJ 6281 数列分块入门 5(分块区间开方区间求和)
    LibreOJ 6280 数列分块入门 4(分块区间加区间求和)
    LibreOJ 6279 数列分块入门 3(分块+排序)
    LibreOJ 6278 数列分块入门 2(分块)
    LibreOJ 6277 数列分块入门 1(分块)
    BZOJ 2301 Problem b(莫比乌斯反演+分块优化)
    MD5Untils加密工具类
    20160418javaweb之 Filter过滤器
    20160417javaweb之servlet监听器
  • 原文地址:https://www.cnblogs.com/claireyuancy/p/7265374.html
Copyright © 2011-2022 走看看