HDU 3001 Travelling

最新推荐文章于 2021-07-15 12:36:48 发布

原创最新推荐文章于 2021-07-15 12:36:48 发布 · 225 阅读

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本文介绍了一个使用三进制状态压缩动态规划解决旅行问题的方法。问题要求遍历所有城市且每个城市最多访问两次，以最小化总费用。通过状态压缩技巧，文章详细展示了如何编码实现这一复杂的旅行路线规划。

After coding so many days,Mr Acmer wants to have a good rest.So travelling is the best choice!He has decided to visit n cities(he insists on seeing all the cities!And he does not mind which city being his start station because superman can bring him to any city at first but only once.), and of course there are m roads here,following a fee as usual.But Mr Acmer gets bored so easily that he doesn't want to visit a city more than twice!And he is so mean that he wants to minimize the total fee!He is lazy you see.So he turns to you for help. Input

There are several test cases,the first line is two intergers n(1<=n<=10) and m,which means he needs to visit n cities and there are m roads he can choose,then m lines follow,each line will include three intergers a,b and c(1<=a,b<=n),means there is a road between a and b and the cost is of course c.Input to the End Of File.

Output

Output the minimum fee that he should pay,or -1 if he can't find such a route.

Sample Input

Sample Output

100
90
7

因为每个点最多走过两次，所以用三进制进行状态压缩。

#include<iostream>//三进制状态压缩DP
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int bit[11]={1,3,9,27,81,243,729,2187,6561,19683,59049};//一共3的10次方个状态
int dp[60000][12];//第一维代表状态，第二维代表当前位置
int num[60000][12];
int map[12][12];
const int INF=0x3f3f3f3f;
int n,m;
int nmin;
void Three()
{
    int i,j,b;
    for(i=0;i<=bit[10]-1;i++)
    {
        b=i;
        for(j=0;j<=9;j++)
        {
            num[i][j]=b%3;
            b/=3;
        }
    }
}
void Init()
{
    int i,j;
    memset(map,-1,sizeof(map));
    for(i=0;i<=bit[n]-1;i++)
    {
        for(j=0;j<=n-1;j++)
        {
            dp[i][j]=INF;
        }
    }
    for(j=0;j<=n-1;j++)
    {
        dp[bit[j]][j]=0;
    }
}
void DP()
{
    int i,j,flag,k,next;
    nmin=INF;
    for(i=0;i<=bit[n]-1;i++)
    {
        flag=1;
        for(j=0;j<=n-1;j++)
        {
            if(num[i][j]==0)
                flag=0;
            if(dp[i][j]==INF)
                continue;
            for(k=0;k<=n-1;k++)
            {
                if(j==k||num[i][k]>=2||map[k][j]==-1)
                    continue;
                next=i+bit[k];
                dp[next][k]=min(dp[next][k],dp[i][j]+map[j][k]);
            }
        }
        if(flag)
        {
            for(j=0;j<=n-1;j++)
            {
                nmin=min(nmin,dp[i][j]);//该状态下的最优解
            }
        }
    }
    return ;
}
int main()
{
    int a,b,c;
    Three();
    while(~scanf("%d%d",&n,&m))
    {
        Init();
        int i,j;
        for(i=0;i<=m-1;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            if(map[a-1][b-1]==-1)
                map[a-1][b-1]=map[b-1][a-1]=c;
            else map[a-1][b-1]=map[b-1][a-1]=min(map[a-1][b-1],c);
        }
        DP();
        if(nmin==INF) nmin=-1;
        printf("%d\n",nmin);
    }
    return 0;
}