3 75 15 21 75 15 28 34 70 5
188
#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<string> #include<algorithm> #include<cstdlib> #include<set> #include<queue> #include<stack> #include<vector> #include<map> #define N 100010 #define Mod 10000007 #define lson l,mid,idx<<1 #define rson mid+1,r,idx<<1|1 #define lc idx<<1 #define rc idx<<1|1 const double EPS = 1e-11; const double PI = acos ( -1.0 ); const double E = 2.718281828; typedef long long ll; const int INF = 1000010; using namespace std; const int M=3030;///测了几个数据,所有情况3000种左右 int a[22][22], cnt[M], n, len; int dp[22][M], sum[22][M]; bool ok ( int x )///判断同一行的x情况是否符合条件,即不能相邻 { if ( x & ( x << 1 ) ) return false; return true; } void build()///将符合条件的数存起来 { len = 0; for ( int i = 0; i < ( 1 << n ); i++ ) if ( ok ( i ) ) cnt[len++] = i; } int getsum ( int i, int x )///计算第i行,x情况时的值 { int sum = 0; int j = 0; while ( x > 0 ) { if ( x & 1 ) sum += a[i][j]; x >>= 1; j++; } return sum; } void finds()///把所有的情况都算出来 { for ( int i = 0; i < n; i++ ) for ( int j = 0; j < len; j++ ) sum[i][j] = getsum ( i, cnt[j] ); } int main() { //freopen("in.txt","r",stdin); while ( cin >> n ) { for ( int i = 0; i < n; i++ ) for ( int j = 0; j < n; j++ ) scanf ( "%d", &a[i][j] ); memset ( dp, -1, sizeof dp ); build(); finds(); for ( int i = 0; i < len; i++ )///先算第0行 dp[0][i] = sum[0][i]; for ( int i = 1; i < n; i++ ) { for ( int j = 0; j < len; j++ ) { int x = -1; for ( int k = 0; k < len; k++ )///找出符合情况的最大值 { if ( dp[i - 1][k] != -1 && ( ! ( cnt[j]&cnt[k] ) ) && dp[i - 1][k] > x ) x = dp[i - 1][k]; } dp[i][j] = x + sum[i][j]; } } int ans = -1; for ( int i = 0; i < len; i++ ) if ( dp[n - 1][i] > ans ) ans = dp[n - 1][i]; cout << ans << endl; } return 0; }
原文地址:http://blog.csdn.net/acm_baihuzi/article/details/40794211