HDU 4960 Another OCD Patient 简单DP

思路：

　　因为是对称的，所以如果两段是对称的，那么一段的前缀和一定等于另一段的后缀和。根据这个性质，我们可以预处理出这个数列的对称点对。然后最后一个对称段是从哪里开始的，做n^2的DP就可以了。

代码：

　　

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <cmath>
 #include <algorithm>
 #include <string>
 #include <queue>
 #include <stack>
 #include <vector>
 #include <map>
 #include <set>
 #include <functional>
 #include <cctype>
 #include <time.h>

 using namespace std;

 typedef __int64 ll;

 const int INF = <<;
 const int MAXN = (int) ;

 inline void nextInt(int &x) {
     char c = getchar();
     x = ;
     while (isdigit(c)) {
         x = x* + c-'';
         c = getchar();
     }
 }

 inline void nextLL(ll &x) {
     char c = getchar();
     x = ;
     while (isdigit(c)) {
         x = x* + c-'';
         c = getchar();
     }
 }

 ll a[MAXN], V[MAXN], prefix[MAXN], suffix[MAXN];
 ll dp[MAXN];
 int sym[MAXN];
 int n;

 void solve() {
     a[] = ;
     prefix[] = suffix[n+] = ;
     for (int i = ; i <= n; i++) prefix[i] = suffix[i] = V[i];
     for (int i = ; i < n; i++) prefix[i+] += prefix[i]; //前缀和
     for (int i = n; i > ; i--) suffix[i] += suffix[i+]; //后缀和

     for (int i = , j = n; i <= n; i++) { //求对称点
         sym[i] = -;
         while (j> && prefix[i]>suffix[j]) j--;
         if (prefix[i]==suffix[j]) sym[i] = j;
     }

     memset(dp, -, sizeof(dp));
     for (int i = ; i <= n; i++) if (sym[i]>) { //这一点有对称点
         if (sym[i] <= i) break; //枚举过界
         dp[i] = a[i] + a[n-sym[i]+]; //前面是一整段
         for (int j = ; j < i; j++) if (sym[j]>) { //从j转移过来
             dp[i] = min(dp[i], dp[j]+a[i-j]+a[sym[j]-sym[i]]);
         }
     }

     ll ans = a[n];
     for (int i = ; i <= n; i++) if (dp[i]>=)
         ans = min(ans, dp[i]+a[sym[i]-i-]); //中间合成一段
     printf("%I64d\n", ans);
 }

 int main() {
     #ifdef Phantom01
         freopen("HDU4960.txt", "r", stdin);
     #endif //Phantom01

     while () {
         nextInt(n);
         if (n==) break;
         for (int i = ; i <= n; i++)
             nextLL(V[i]);
         for (int i = ; i <= n; i++)
             nextLL(a[i]);
         solve();
     }

     return ;
 }