【洛谷习题】售货员的难题

题目链接：https://www.luogu.org/problemnew/show/P1171

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比较经典的状压DP吧，就是最后一个点不开O2卡不过去。。。

主要是对位运算熟，注意f[1][0]=0，其他状态为正无穷，别的看代码吧。

#include <cstdio>
#include <cstring>

using namespace std;

inline int get_num() {
    int num = 0;
    char c = getchar();
    while (c < '0' || c > '9') c = getchar();
    while (c >= '0' && c <= '9')
        num = num * 10 + c - '0', c = getchar();
    return num;
}

const int maxn = 20;

int dist[maxn][maxn], f[1 << maxn][maxn];

int main() {
    int n, ans = 0x3f3f3f3f;
    n = get_num();
    for (register int i = 0; i < n; ++i)
        for (register int j = 0; j < n; ++j)
            dist[i][j] = get_num();
    memset(f, 0x3f, sizeof(f));
    f[1][0] = 0;
    for (register int i = 1; i < 1 << n; ++i)
        for (register int j = 0; j < n; ++j) if (i >> j & 1)
            for (register int k = 0; k < n; ++k) if ((i ^ 1 << j) >> k & 1)
                if (f[i ^ 1 << j][k] + dist[k][j] < f[i][j])
                    f[i][j] = f[i ^ 1 << j][k] + dist[k][j];
    for (register int i = 0; i < n; ++i)
        if (f[(1 << n) - 1][i] + dist[i][0] < ans)
            ans = f[(1 << n) - 1][i] + dist[i][0];
    printf("%d", ans);
    return 0;
}