SGU 482 Impudent Thief dp

来源：互联网发布：王宝强人品知乎编辑：程序博客网时间：2024/06/02 02:16

题目链接：点击打开链接

题意：

给定n列箱子，下面n个数表示每列的箱子个数

每次可以拿走一列箱子，拿走中间列的箱子后两边的箱子会向中间靠拢（即中间不会有缝隙）

拿走箱子后的周长必须 *2 >= 原周长

问：最多拿走几个箱子（即使得面积最大）

输出能拿走的最大面积

输出拿走的列数

输出拿走具体哪几列

思路：

dp[i][j]表示前i列留下第i列后剩下面积为j时最大的周长。

顺便记录下前驱什么的。。

#include<iostream>#include<stdio.h>#include<string.h>#include <algorithm>#include<string>#include<set>#include<vector>#include<queue>#include<math.h>template <class T>inline bool rd(T &ret) {char c; int sgn;if (c = getchar(), c == EOF) return 0;while (c != '-' && (c<'0' || c>'9')) c = getchar();sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');ret *= sgn;return 1;}template <class T>inline void pt(T x) {if (x <0) {putchar('-');x = -x;}if (x>9) pt(x / 10);putchar(x % 10 + '0');}using namespace std;typedef long long ll;typedef pair<int, int> pii;const int N = 51;int a[N], n;int dp[N][5005];pii pre[N][5005];int go(int x, int y){if (x >= y)return 2;else return (y - x + 1)*2;}set<int>G;vector<int> D;void dfs(pii x){if (x.first <= 0)return;G.insert(x.first);x = pre[x.first][x.second];dfs(x);}int main(){while (cin >> n){ int all = 0, len = 0;for (int i = 1; i <= n; i++){rd(a[i]);all += a[i];len += go(a[i - 1], a[i]);}memset(dp, -1, sizeof dp);for (int i = 0; i <= n; i++)for (int j = 0; j <= i * 100; j++)pre[i][j] = pii(-1, -1);dp[0][0] = 0;for (int i = 0; i < n; i++)for (int j = 0; j <= i * 100; j++){if (dp[i][j] == -1)continue;for (int k = i + 1; k <= n; k++){if (dp[k][j + a[k]] >= dp[i][j] + go(a[i], a[k]))continue;dp[k][j + a[k]] = dp[i][j] + go(a[i], a[k]);pre[k][j + a[k]] = pii(i, j);}}len = (len + 1) >> 1;int ans = 0; pii lala;for (int i = 1; i <= n; i++)for (int j = 0; j <= i * 100; j++)if (dp[i][j] >= len && ans <= all - j){ans = all - j;lala = pii(i, j);}pt(ans); puts("");G.clear();dfs(lala);D.clear();for (int i = 1; i <= n; i++)if (!G.count(i))D.push_back(i);pt((int)D.size()); puts("");for (int i = 0; i < D.size(); i++)printf("%d%c", D[i], i + 1 == D.size() ? '\n' : ' ');}return 0;}

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