HDUOJ 3001 - Travelling

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 

 

题意是，要遍历所有的城市，不过一个城市最多走两遍，这样可以用3进制来表示，可以想到用状态压缩。

ans[i][j]表示的是在点集i的情况下，以j为起点。

遍历所有的3进制的情况，一开始先将所有的【在这个题目范围内】数字全都转换成3进制进行保存，这样用的时候就可以直接查询了。

具体提示看注释。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int INF = (1<<30) - 1;
int maps[12][12];
int bit[11] = {1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049};
int ans[60000][12];///ans[i][j]表示在i状态下去j的最小值
int num_tri[60000][12];///记录每个数的三进制的数

int main()
{
    int n, m;
    int minn;
    for (int i = 0; i < bit[10]; ++i)///计算所有数字的三进制数
    {
        int key = i;
        for (int j = 0; j < 10; ++j)
        {
            num_tri[i][j] = key%3;
            key /= 3;
        }
    }
    while (scanf("%d%d", &n, &m) != EOF)
    {
        memset(maps, -1, sizeof(maps));
        for (int i = 0; i < bit[n]; ++i)///将所有的路径初始化为无穷大
        {
            for (int j = 0; j < n; ++j)
                ans[i][j] = INF;
        }
        for (int i = 0; i < n; ++i)///从这个点到他自己设为0
            ans[ bit[i] ][i] = 0;

        for (int i = 0; i < m; ++i)
        {
            int from, to, v;
            scanf("%d%d%d", &from, &to, &v);
            if (maps[from-1][to-1] == -1)
                maps[from-1][to-1] = maps[to-1][from-1] = v;
            else
                maps[from-1][to-1] = maps[to-1][from-1] = min(maps[from-1][to-1], v);
        }
        int flag;
        minn = INF;
        for (int i = 0; i < bit[n]; ++i)
        {
            flag = 1;
            for (int j = 0; j < n; ++j)
            {
                if (num_tri[i][j] == 0)///如果在i点集的情况下没有去j
                    flag = 0;
                if (ans[i][j] == INF)///如果之前没有初始化过这个点
                    continue;
                for (int k = 0; k < n; ++k)///判断以哪个点为起点
                {
                    if (k == j || num_tri[i][k] >= 2 || maps[k][j] == -1)
                        continue;
                    int next = i + bit[k];
                    ans[next][k] = min(ans[next][k], ans[i][j] + maps[j][k]);
                    //printf("DP Start\n");
                }
            }
            if (flag == 1)
            {
                for (int j = 0; j < n; ++j)
                    minn = min(minn, ans[i][j]);
            }
        }
        if (minn == INF)
            minn = -1;
        printf("%d\n", minn);
    }
    return 0;
}