PKU 3311 Hie with the Pie 状态DP

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int INF = (int) 1e8;

int dp[1<<11][11]; //dp[S][v] 表示还需访问的集合为S, 现在在v点
int d[11][11];
int n;

void 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]);
    }
}

void print() {
    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= n; j++)
            printf("%d ", d[i][j]);
        puts("");
    }
}

int main() {
    #ifdef Phantom01
        freopen("PKU3311.txt", "r", stdin);
    #endif // Phantom01

    while (scanf("%d", &n)!=EOF) {
        if (0==n) break;
        for (int i = 0; i <= n; i++)
            for (int j = 0; j <= n; j++)
                scanf("%d", &d[i][j]);
//        print();
        floyd();
//        print();

        for (int i = 0; i <= n; i++)
            for (int S = 0; S < (1<<(n+1)); S++)
                dp[S][i] = INF;

        dp[(1<<(n+1))-1][0] = 0;
        for (int S = (1<<(n+1))-2; S >= 0; S--) {
            for (int v = 0; v <= n; v++) {
                for (int u = 0; u <= n; u++) {
                    if (!(S&(1<<u))) {
                        if (dp[S][v] < 0) {
                            dp[S][v] = dp[S|(1<<u)][u] + d[v][u];
                        } else {
                            dp[S][v] = min(dp[S][v], dp[S|(1<<u)][u] + d[v][u]);
                        }
                    }
                }
            }
        }
        printf("%d\n", dp[0][0]);
    }

    return 0;
}