POJ 3123

https://cn.vjudge.net/problem/POJ-3123

題意：

n 個城市，m 條路，給定八個點（也就是四對），使每隊點連通且總權和最小。

分析：

dp[i][j] 表示 i 狀態下以 j 爲起點的最小總權和。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <bitset>
#include <map>
typedef long long int ll;
const int MOD = (int)1e9 + 7;
const int INF = 99999999;
using namespace std;

int n, m;
map<string, int> city;
int dp1[1 << 8][35];
int dp2[1 << 8];
int d[35][35];
int vis[35];

bool check(int s)
{
    for (int i = 0; i < 4; i++)
    {
        if (((s & 3) != 3) && ((s & 3) != 0))
            return false;

        s >>= 2;
    }

    return true;
}

int main()
{
    while (~scanf("%d%d", &n, &m) && (n || m))
    {
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                d[i][j] = (i == j) ? 0 : INF;

        string s1, s2;
        int w;
        for (int i = 0; i < n; i++)
        {
            cin >> s1;
            city[s1] = i;
        }
        for (int i = 0; i < m; i++)
        {
            cin >> s1 >> s2 >> w;
            int u = city[s1];
            int v = city[s2];

            if (w < d[u][v])
                d[u][v] = d[v][u] = w;
        }

        // Floyd
        for (int k = 0; k < n; k++)
            for (int i = 0; i < n; i++)
                for (int j = 0; j < n; j++)
                    d[i][j] = min(d[i][j], d[i][k] + d[k][j]);

        for (int i = 0; i < 8; i++)
        {
            cin >> s1;

            for (int j = 0; j < n; j++)
                dp1[1 << i][j] = d[j][city[s1]];
        }

        for (int i = 0; i < (1 << 8); i++)
        {
            if (!(i & (i - 1)))
                continue;

            for (int j = 0; j < n; j++)
            {
                dp1[i][j] = INF;

                for (int sub = (i - 1) & i; sub != 0; sub = (sub - 1) & i)
                    dp1[i][j] = min(dp1[i][j], dp1[sub][j] + dp1[i - sub][j]);
            }

            memset(vis, 0, sizeof(vis));

            int min_w, min_i;
            for (int j = 0; j < n; j++)
            {
                min_w = INF;

                for (int k = 0; k < n; k++)
                {
                    if (dp1[i][k] < min_w && !vis[k])
                    {
                        min_w = dp1[i][k];
                        min_i = k;
                    }
                }

                vis[min_i] = 1;

                for (int k = 0; k < n; k++)
                    dp1[i][min_i] = min(dp1[i][min_i], dp1[i][k] + d[k][min_i]);
            }
        }

        for (int i = 0; i < (1 << 8); i++)
        {
            dp2[i] = INF;

            for (int j = 0; j < n; j++)
                dp2[i] = min(dp2[i], dp1[i][j]);
        }

        for (int i = 0; i < (1 << 8); i++)
        {
            if (check(i))
            {
                for (int j = i; j != 0; j = (j - 1) & i)
                {
                    if (check(j))
                        dp2[i] = min(dp2[i], dp2[j] + dp2[i - j]);
                }
            }
        }

        printf("%d\n", dp2[(1 << 8) - 1]);
    }
    return 0;
}