HDU 3001 Travelling(狀態壓縮dp)

Travelling

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5765    Accepted Submission(s): 1858

Problem Description

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

2 1 1 2 100 3 2 1 2 40 2 3 50 3 3 1 2 3 1 3 4 2 3 10

 

Sample Output

100 90 7

 

Source

2009 Multi-University Training Contest 11 - Host by HRBEU

 

題意:要走完所有的城市，每條邊有一個代價，求走完所有的城市的最小代價，起初沒看到每個城市最多可走兩邊。

思路:用二進制的狀態壓縮果然wa了，用三進制的狀態壓縮，先把進製表打好，然後用一個二維的數組表示出這個數每位在三進制下的狀態，狀態轉移。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<map>
#include<cstring>
#include<map>
#include<set>
#include<queue>
#include<stack>
using namespace std;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f, N = 6e4;
typedef long long ll;
int dp[N][12], Map[12][12], n, m, three[12], vis[N][12];
void init()
{
    three[0] = 1;
    for(int i = 1; i<=10; i++)
        three[i] = three[i-1] * 3;
    for(int i = 0; i<three[10]; i++)
    {
        int temp = i;
        for(int j = 1; j<=10; j++)
        {
            vis[i][j] = temp % 3;
            temp /= 3;
        }
    }
}
int main()
{
    init();
    while(~scanf("%d%d", &n, &m))
    {
        memset(Map, 0x3f, sizeof(Map));
        int u, v, w;
        while(m--)
        {
            scanf("%d%d%d", &u, &v, &w);
            Map[u][v] = Map[v][u] = min(w, Map[u][v]);
        }
        memset(dp, 0x3f, sizeof(dp));
        int num = three[n] - 1;
        for(int i = 1; i<=n; i++)
            dp[three[i-1]][i] = 0;
        int ans = inf;
        for(int i = 1; i<=num; i++)
        {
            for(int j = 1; j<=n; j++)
            {
                if(dp[i][j] == inf)
                    continue;
                for(int k = 1; k<=n; k++)
                {
                    if(vis[i][k]>=2 || Map[j][k] == inf || j == k)
                        continue;
                    dp[i + three[k-1]][k] = min(dp[i + three[k-1]][k], dp[i][j] + Map[j][k]);

                }
            }
        }
        for(int i = 1; i<=num; i++)
        {
            int ok = 1;
            for(int j = 1; j<=n; j++)
                if(!vis[i][j])
                {
                    ok  = 0;
                    break;
                }
            if(ok)
                for(int j = 1; j<=n; j++)
                    ans = min(dp[i][j], ans);
        }
        if(ans != inf)
            cout<<ans<<endl;
        else cout<<-1<<endl;
    }
    return 0;
}