HZAU 1199: Little Red Riding Hood 01背包

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题目链接：1199: Little Red Riding Hood

思路：dp(i)表示前i朵花能取得的最大价值，每一朵花有两种选择，摘与不摘，摘了第i朵花后第i-k到i+k的花全部枯萎，那么摘的话dp(i) = dp(i-k-1) + a[i]，不摘就是dp(i) = dp(i-1)，因此转移方程就是dp(i) = max(dp(i-k-1) + a[i], dp(i-1))。

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 = 1e5 + 5;int a[maxn], dp[maxn];int main() {int T, n, k;scanf("%d", &T);while(T--) {scanf("%d%d", &n, &k);for(int i = 1; i <= n; ++i) {scanf("%d", &a[i]);}dp[0] = 0;for(int i = 1; i <= n; ++i) {dp[i] = dp[i-1];int w;if(i - k - 1 <= 0) w = 0;else w = dp[i-k-1];dp[i] = max(dp[i], w + a[i]);}printf("%d\n", dp[n]);}return 0;}

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