HDU 4960 Another OCD Patient(记忆化搜索)

时间:2022-10-11 23:52:11

HDU 4960 Another OCD Patient

pid=4960" target="_blank" style="">题目链接

记忆化搜索,因为每一个碎片值都是正数,所以每一个前缀和后缀都是递增的,就能够利用twopointer去找到每一个相等的位置,然后下一个区间相当于一个子问题,用记忆化搜索就可以,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;
}