HDU 4960 Another OCD Patient 簡單DP

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思路：

　　因為是對稱的，所以如果兩段是對稱的，那麼一段的首碼和一定等於另一段的尾碼和。根據這個性質，我們可以預先處理出這個數列的對稱點對。然後最後一個對稱段是從哪裡開始的，做n^2的DP就可以了。

 

代碼：

　　

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <cmath> 6 #include <algorithm> 7 #include <string> 8 #include <queue> 9 #include <stack>10 #include <vector>11 #include <map>12 #include <set>13 #include <functional>14 #include <cctype>15 #include <time.h>16 17 using namespace std;18 19 typedef __int64 ll;20 21 const int INF = 1<<30;22 const int MAXN = (int) 5055;23 24 inline void nextInt(int &x) {25     char c = getchar();26     x = 0;27     while (isdigit(c)) {28         x = x*10 + c-‘0‘;29         c = getchar();30     }31 }32 33 inline void nextLL(ll &x) {34     char c = getchar();35     x = 0;36     while (isdigit(c)) {37         x = x*10 + c-‘0‘;38         c = getchar();39     }40 }41 42 ll a[MAXN], V[MAXN], prefix[MAXN], suffix[MAXN];43 ll dp[MAXN];44 int sym[MAXN];45 int n;46 47 void solve() {48     a[0] = 0;49     prefix[0] = suffix[n+1] = 0;50     for (int i = 1; i <= n; i++) prefix[i] = suffix[i] = V[i];51     for (int i = 0; i < n; i++) prefix[i+1] += prefix[i]; //首碼和52     for (int i = n; i > 0; i--) suffix[i] += suffix[i+1]; //尾碼和53 54     for (int i = 1, j = n; i <= n; i++) { //求對稱點55         sym[i] = -1;56         while (j>0 && prefix[i]>suffix[j]) j--;57         if (prefix[i]==suffix[j]) sym[i] = j;58     }59 60     memset(dp, -1, sizeof(dp));61     for (int i = 1; i <= n; i++) if (sym[i]>0) { //這一點有對稱點62         if (sym[i] <= i) break; //枚舉過界63         dp[i] = a[i] + a[n-sym[i]+1]; //前面是一整段64         for (int j = 1; j < i; j++) if (sym[j]>0) { //從j轉移過來65             dp[i] = min(dp[i], dp[j]+a[i-j]+a[sym[j]-sym[i]]);66         }67     }68 69     ll ans = a[n];70     for (int i = 1; i <= n; i++) if (dp[i]>=0)71         ans = min(ans, dp[i]+a[sym[i]-i-1]); //中間合成一段72     printf("%I64d\n", ans);73 }74 75 int main() {76     #ifdef Phantom0177         freopen("HDU4960.txt", "r", stdin);78     #endif //Phantom0179 80     while (1) {81         nextInt(n);82         if (n==0) break;83         for (int i = 1; i <= n; i++)84             nextLL(V[i]);85         for (int i = 1; i <= n; i++)86             nextLL(a[i]);87         solve();88     }89 90     return 0;91 }

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