ZOJ 2059 The Twin Towers
Posted Fighting Heart
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了ZOJ 2059 The Twin Towers相关的知识,希望对你有一定的参考价值。
双塔DP。
dp[i][j]表示前i个物品,分成两堆(可以不全用),价值之差为j的时候,较小一堆的价值为dp[i][j]。
#include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; int dp[105][4000 + 10]; int a[105]; int n, sum; void read() { for (int i = 1; i <= n; i++) scanf("%d", &a[i]); } void work() { sum = 0; for (int i = 1; i <= n; i++) sum = sum + a[i]; memset(dp, -1, sizeof dp); int h; h = a[1]; dp[1][h + sum] = dp[1][0 - h + sum] = dp[1][sum] = 0; for (int i = 2; i <= n; i++) { h = a[i]; for (int j = 0; j <= sum*2; j++) dp[i][j] = dp[i - 1][j]; for (int j = 0; j <= sum*2; j++) { if (dp[i - 1][j] == -1) continue; int tmp = j - sum; if (tmp >= 0) { dp[i][h + tmp + sum] = max(dp[i][h + tmp + sum], dp[i - 1][j]); dp[i][tmp - h + sum] = max(dp[i][tmp - h + sum], dp[i - 1][j] + min(tmp, h)); } else if (tmp<0) { tmp = -tmp; dp[i][0 - (tmp + h) + sum] = max(dp[i][0 - (tmp + h) + sum], dp[i - 1][j]); dp[i][h - tmp + sum] = max(dp[i][h - tmp + sum], dp[i - 1][j] + min(tmp, h)); } } } int ans = 0; for (int i = 1; i <= n; i++) ans = max(ans, dp[i][sum]); if (ans == 0) printf("Sorry\n"); else printf("%d\n", ans); } int main() { while (~scanf("%d", &n)) { read(); if (n < 0) break; if (n <=1) printf("Sorry\n"); else work(); } return 0; }
以上是关于ZOJ 2059 The Twin Towers的主要内容,如果未能解决你的问题,请参考以下文章
[zoj] 3229 Shoot the Bullet || 有源汇上下界最大流
代写HTML5 JavascriptWEB作业代写JSPASPAsp.net作业A simple animation Towers of Hanoi
The Towers of Hanoi Revisited---(多柱汉诺塔)
Acdream 1219 The Towers of Hanoi Revisited(递归汉诺塔问题)
Navicate连接MySQL出现2059-Authentication plugin 'caching_sha2_password' cannot be loaded:的解决方案(代
SQLyog连接MySQL8 异常2059-Authentication plugin 'caching_sha2_password' cannot be loaded解决方案(示例代