HDU - 4960 Another OCD Patient 记忆化

题意：

给定序列a，合并多个连续的段，使得整个序列成为回文序列；下一行给定序列v，v[i]表示合并i个数需要的花费；

思路：

可以直接记忆化往下递归，可能数据不够强，能够（900ms）卡着时间过，当然内存消耗也很大；

另一种真实的dp见下篇博客； 手动滑稽

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn = 5000 + 7;
const ll mod = 1e9+7;

int n;
ll a[maxn], v[maxn], sum[maxn];
bool vis[maxn][maxn];
ll dp[maxn][maxn];

int find_(int l, int r, ll v) {
    int l_ = l, r_ = r;
//    int res = 0;
    while(l_ <= r_) {
        int mid = (l_+r_) / 2;
        if(sum[r]-sum[mid-1] == v) return mid;
        if(sum[r]-sum[mid-1] > v) l_ = mid+1;
        else r_ = mid-1;
    }
    return 0;
}

ll dfs(int l, int r) {
    if(l >= r) return 0;
    if(vis[l][r]) return dp[l][r];
    vis[l][r] = 1;
    ll ans = v[r-l+1];
    for(int i = l; i < r; ++i) {
        int r_ = find_(i+1, r, sum[i]-sum[l-1]);
        if(r_ == 0) continue;
        ans = min(ans, dfs(i+1, r_-1)+v[i-l+1]+v[r-r_+1]);
    }
    return dp[l][r] = ans;
}

int main() {
    while(~scanf("%d", &n) && n) {
        memset(vis, 0, sizeof vis);
        for(int i = 1; i <= n; ++i) {
            scanf("%lld", &a[i]);
            sum[i] = sum[i-1] + a[i];
        }
        for(int i = 1; i <= n; ++i) {
            scanf("%lld", &v[i]);
        }
        printf("%lld\n", dfs(1, n));
    }

    return 0;
}