HDU 4960 Another OCD Patient

题目链接

记忆化搜索，由于每个碎片值都是正数，所以每个前缀和后缀都是递增的，就可以利用twopointer去找到每个相等的位置，然后下一个区间相当于一个子问题，用记忆化搜索即可，复杂度接近O(n^2)

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 5005;

typedef long long ll;

int n, a[N], dp[N][N];
ll v[N], pre[N];

void init() {
    for (int i = 1; i <= n; i++) {
	scanf("%I64d", &v[i]);
	pre[i] = pre[i - 1] + v[i];
    }
    for (int i = 1; i <= n; i++)
	scanf("%d", &a[i]);
    memset(dp, -1, sizeof(dp));
}

int solve(int l, int r) {
    if (dp[l][r] != -1) return dp[l][r];
    dp[l][r] = a[r - l + 1];
    if (l >= r) return dp[l][r] = 0;
    int now = l;
    for (int i = r; i >= l; i--) {
	while (pre[now] - pre[l - 1] < pre[r] - pre[i - 1] && now < i)
	    now++;
	if (now == i) break;
	if (pre[now] - pre[l - 1] == pre[r] - pre[i - 1])
	    dp[l][r] = min(dp[l][r], a[now - l + 1] + a[r - i + 1] + solve(now + 1, i - 1));
    }
    return dp[l][r];
}

int main() {
    while (~scanf("%d", &n) && n) {
	init();
	printf("%d\n", solve(1, n));
    }
    return 0;
}