HDU1565——方格取数(1)

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方格取数(1)

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5726 Accepted Submission(s): 2167

Problem Description

给你一个n*n的格子的棋盘，每个格子里面有一个非负数。
从中取出若干个数，使得任意的两个数所在的格子没有公共边，就是说所取的数所在的2个格子不能相邻，并且取出的数的和最大。

Input

包括多个测试实例，每个测试实例包括一个整数n 和n*n个非负数(n<=20)

Output

对于每个测试实例，输出可能取得的最大的和

Sample Input

375 15 21 75 15 28 34 70 5

Sample Output

Author

ailyanlu

Source

Happy 2007

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状态压缩，先dfs预处理出每一行可能的所有状态，设dp[i][j] 表示第i行状态为j时，最大可以达到的数

dp[i][j] = max(dp[i - 1][k]) + sum;
sum为状态j对应可以取到的数的和

#include <map>  #include <set>  #include <list>  #include <stack>  #include <vector>  #include <queue>  #include <cmath>  #include <cstdio>  #include <cstring>  #include <iostream>  #include <algorithm>  using namespace std;int n, res;int sta[4040];int dp[22][4040];int mat[22][22];void dfs(int l, int status){if (l == n){sta[res++] = status;return ;}dfs(l + 1, status << 1);if ( !(status & 1) ){dfs(l + 1, status << 1 | 1);}}int main(){while (~scanf("%d", &n)){res = 0;dfs(0, 0);for (int i = 1; i <= n; ++i){for (int j = 1; j <= n; ++j){scanf("%d", &mat[i][j]);}}memset (dp, 0, sizeof(dp) );for (int i = 1; i <= n; ++i){for (int j = 0; j < res; ++j){int tmp = 0;for (int k = 0; k < res; ++k){if ( !(sta[j] & sta[k]) ){tmp = max(tmp, dp[i - 1][k]);}}int sum = 0;for (int l = 0; l < n; ++l){if( sta[j] & (1 << l) ){sum += mat[i][n - l];}}dp[i][j] = tmp + sum;}}int ans = 0;for (int i = 0; i < res; ++i){ans = max(ans, dp[n][i]);}printf("%d\n", ans);}return 0;}

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