hihoCoder 1044 : 状态压缩·一 状压dp

              思路：状态压缩，dp(i, j)表示考虑前i个数且[i-m+1, i]的选择情况为j。如果要选择当前这个数并且，数位1的个数不超过q，则dp[i+1][nex] = max(dp[i+1][nex], dp[i][j] + w[i+1])，如果不选择dp[i+1][j>>1] = max(dp[i+1][j>>1], dp[i][j]);

AC代码

#include <cstdio>#include <cmath>#include <cctype>#include <algorithm>#include <cstring>#include <utility>#include <string>#include <iostream>#include <map>#include <set>#include <vector>#include <queue>#include <stack>using namespace std;#pragma comment(linker, "/STACK:1024000000,1024000000") #define eps 1e-10#define inf 0x3f3f3f3f#define PI pair<int, int> typedef long long LL;const int maxn = 1000 + 5;int dp[maxn][1<<10], w[maxn];int main() {int n, m, q;while(scanf("%d%d%d", &n, &m, &q) == 3){for(int i = 1; i <= n; ++i) {scanf("%d", &w[i]);}memset(dp, 0, sizeof(dp));int tol = 1<<m;for(int i = 0; i < n; ++i) {for(int j = 0; j < tol; ++j) {int nex = (j>>1)^(1<<(m-1));int cnt = 0;for(int k = 0; k < m; ++k) {if(nex & (1<<k)) ++cnt;}if(cnt <= q) dp[i+1][nex] = max(dp[i+1][nex], dp[i][j] + w[i+1]);dp[i+1][j>>1] = max(dp[i+1][j>>1], dp[i][j]);}}int ans = 0;for(int i = 0; i < tol; ++i) {ans = max(ans, dp[n][i]);}printf("%d\n", ans);}return 0;} 

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