nyoj234 吃土豆 01背包

来源：互联网发布：惠普游戏本知乎编辑：程序博客网时间：2024/04/30 06:37

思路：假设我们先只考虑一行，规则就是取了i处的土豆，每一个土豆有两种选择，拿与不拿，那么i-1和i+1处的土豆都不能再取，那么要求某一行的最大取值就用一次动态规划即可，dp(i)表示前i个土豆能取得的最大值，转移方程dp(i) = max{dp(i-1), dp(i-2) + w[i]}，此时我们已经得到了每一行的最优解，然后可以把n行看做是n个土豆，即每一行看做最优值土豆，同样的状态方程。此题很妙。

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 = 500 + 5;int w[maxn][maxn], sum[maxn][maxn], dp[maxn];int main() {int n, m;while(scanf("%d%d", &n, &m) == 2) {for(int i = 0; i < n; ++i) {for(int j = 0; j < m; ++j) {scanf("%d", &w[i][j]);sum[i][j] = j==0?0:sum[i][j-1];int pre;if(j-2 < 0) pre = 0;else pre = sum[i][j-2];sum[i][j] = max(sum[i][j], pre + w[i][j]);}}for(int i = 0; i < n; ++i) {dp[i] = i == 0 ? 0 : dp[i-1];int pre;if(i-2 < 0) pre = 0;else pre = dp[i-2]; dp[i] = max(dp[i], pre + sum[i][m-1]);}printf("%d\n", dp[n-1]);}return 0;}

如有不当之处欢迎指出！

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