HDU 4960 Another OCD Patient (memory-based search), hdu4960

HDU 4960 Another OCD Patient (memory-based search), hdu4960
HDU 4960 Another OCD Patient

Question Link

Memory-based search, because each shard value is a positive number, so each prefix and suffix are incremented, you can use twopointer to find each equal position, and then the next interval is equivalent to a subproblem, it can be searched with memory, and the complexity is close to O (n ^ 2)

Code:

#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;}