HDU 4960 (区间DP)

题目链接：点击这里

题意：把一个数列合并成回文序列，每一个数字只能和并一次，只能连续的子串进行合并，合并后的数字是他们的和，花费是他们中数字个数的函数。给出原始串和花费函数求最小的花费。

O(n)扫一遍相等的前缀的后缀，dpi表示i下标和对应下标的后缀合并之后的最小花费，那么dpi=min{dpj+cost(i,j−1)∥∥i≤j}

#include <bits/stdc++.h>
using namespace std;
#define maxn 5005
int n;
int a[maxn];
int v[maxn];
int x[maxn], y[maxn];
long long dp[maxn];
long long sumx[maxn];
long long sumy[maxn];
int main () {
//freopen("1.txt","r",stdin);
while(scanf("%d",&n)&&(n!=0)) {
sumx[0] = 0, sumy[n+1] = 0;
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]), sumx[i] = sumx[i-1]+a[i];
for (int i = n; i >= 1; i--) sumy[i] = sumy[i+1]+a[i];
for (int i = 1; i <= n; i++)
scanf ("%d", &v[i]);
int L = 1, R = n, tt = 0;
while (L <= R+1) {
if (sumx[L-1] == sumy[R+1]) {
x[++tt] = L, y[tt] = R;
L++, R--;
}
while (L <= R+1 && sumx[L-1] < sumy[R+1]) L++;
while (L <= R+1 && sumx[L-1] > sumy[R+1]) R--;
}

for (int i = tt; i >= 1; i--) {
dp[i] = v[y[i]-x[i]+1]; //不断开
for (int j = i+1; j <= tt; j++) {
int num_l = x[j]-x[i], num_r = y[i]-y[j];
dp[i] = min (dp[i], dp[j]+v[num_l]+v[num_r]);
}
}
printf ("%lld\n", dp[1]);
}
return 0;
}