vijos1037 双塔问题

题意:n个任意高度的水晶,建造两座塔,选取任意快,使得塔尽量高,且塔相等

思路:
分情况说：第i快石头

1、不用：dp[i-1][j]

2、放在高塔上：dp[i-1][j-a[i]]+a[i]

3、放在低塔上并且放后不超过高塔：dp[i-1][j+a[i]]

4、放在低塔上超过了高塔：dp[i-1][a[i]-j]+j
代码:

#include <bits/stdc++.h>
#define LL long long
#define INF 0x3f3f3f3f
#define maxn 2004
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
using namespace std;
int dp[102][maxn];
int h[maxn];
int n, T;
int f, t;
int main()
{
    mem(dp, 0);
    scanf("%d", &n);
    int sum = 0;
    //for (int i = 0; i <= n; ++i)
    for (int j = 1; j < maxn; ++j) dp[0][j] = -INF;
    for (int i = 1; i <= n; ++i) {
        scanf("%d", &h[i]);
        sum += h[i];
    }
    for (int i = 1; i <= n; ++i) {
        for (int j = 0; j <= 2000; ++j) {
            dp[i][j] = max(dp[i-1][j], dp[i-1][j+h[i]]);//不用：dp[i-1][j] 和 放在低塔上并且放后不超过高塔：dp[i-1][j+a[i]]
            if(j >= h[i]) dp[i][j] = max(dp[i][j], dp[i - 1][j - h[i]] + h[i]);//放在高塔上：dp[i-1][j-a[i]]+a[i]
            else dp[i][j] = max(dp[i][j], dp[i - 1][h[i] - j] + j);//放在低塔上超过了高塔：dp[i-1][a[i]-j]+j
        }
    }
    if(dp[n][0] > 0) cout << dp[n][0] << endl;
    else cout << "Impossible\n" << endl;
    return 0;
}