LightOJ 1011 - Marriage Ceremonies(状压DP)

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题意：n男n女，第i男与第j女结合的权值为a[i][j]，求最大权值和。

思路：状压dp。dp[i][s]记录前i个男生与状态数为s表示的i个女生结合的最大权值和。转移方程：dp[i][s|(1 << j)] = max(dp[i][s|(1 << j)], dp[i-1][s] + a[i][j])。

#include <algorithm>#include <iostream>#include <sstream>#include <cstring>#include <cstdio>#include <vector>#include <string>#include <queue>#include <stack>#include <cmath>#include <set>#include <map>using namespace std;typedef long long LL;#define mem(a, n) memset(a, n, sizeof(a))#define rep(i, n) for(int i = 0; i < (n); i ++)#define REP(i, t, n) for(int i = (t); i < (n); i ++)#define FOR(i, t, n) for(int i = (t); i <= (n); i ++)#define ALL(v) v.begin(), v.end()#define si(a) scanf("%d", &a)#define sii(a, b) scanf("%d%d", &a, &b)#define siii(a, b, c) scanf("%d%d%d", &a, &b, &c)#define pb push_back#define eps 1e-8const int inf = 0x3f3f3f3f, N = 17, MOD = 1e9 + 7;int T, cas = 0;int n, m;int a[N][N], dp[N][1 << N];int main(){#ifdef LOCAL    freopen("/Users/apple/input.txt", "r", stdin);//freopen("/Users/apple/out.txt", "w", stdout);#endif    si(T);    while(T --) {    si(n);    for(int i = 0; i < n; i ++)     for(int j = 0; j < n; j ++)     si(a[i][j]);    mem(dp, 0);    for(int i = 0; i < n; i ++) dp[0][1 << i] = a[0][i];    int all = (1 << n) - 1;    for(int i = 1; i < n; i ++) {    for(int s = 0; s <= all; s ++) {    if(!dp[i-1][s]) continue;    for(int j = 0; j < n; j ++) {    if(s & (1 << j)) continue;    dp[i][s|(1 << j)] = max(dp[i][s|(1 << j)], dp[i-1][s] + a[i][j]);    }    }    }    printf("Case %d: %d\n", ++ cas, dp[n-1][all]);    }        return 0;}

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