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：

 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]; //前缀和
52     for (int i = n; i > 0; i--) suffix[i] += suffix[i+1]; //后缀和
53
54     for (int i = 1, j = n; i <= n; i++) { //求对称点
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) { //这一点有对称点
62         if (sym[i] <= i) break; //枚举过界
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) { //从j转移过来
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]); //中间合成一段
72     printf("%I64d\n", ans);
73 }
74
75 int main() {
76     #ifdef Phantom01
77         freopen("HDU4960.txt", "r", stdin);
78     #endif //Phantom01
79
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 }

HDU 4960 Another OCD Patient 简单DP,布布扣,bubuko.com