P1880-[NOI1995]石子合并 

 1 #include <bits/stdc++.h>
 2 #define _for(i,a,b) for(int i = (a);i < b;i ++)
 3 typedef long long ll;
 4 using namespace std;
 5 inline ll read()
 6 {
 7     ll ans = 0;
 8     char ch = getchar(), last = ' ';
 9     while(!isdigit(ch)) last = ch, ch = getchar();
10     while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
11     if(last == '-') ans = -ans;
12     return ans;
13 }
14 inline void write(ll x)
15 {
16     if(x < 0) x = -x, putchar('-');
17     if(x >= 10) write(x / 10);
18     putchar(x % 10 + '0');
19 }
20 int N;
21 int a[309];
22 int dp[309][309];
23 int dp2[309][309];
24 int pre[309];
25 
26 int main()
27 {
28     N = read();
29     _for(i,0,N)
30         a[i] = a[i+N] = read();
31     
32     memset(dp,0,sizeof(dp));
33     _for(i,0,2*N-1)
34         dp[i][i+1] = a[i]+a[i+1];
35         
36     memset(dp2,0,sizeof(dp2));
37     _for(i,0,2*N-1)
38         dp2[i][i+1] = a[i]+a[i+1];
39     
40     pre[0] = a[0];
41     _for(i,1,2*N)
42         pre[i] = pre[i-1]+a[i];
43 
44     for(int d = 3;d < N+1;d ++)
45     {
46         for(int i = 0;i < 2*N-d+1;i ++)
47         {
48             int j = i+d-1;
49             dp2[i][j] = 0x3f3f3f3f;
50             for(int k = i;k < j;k ++)
51             {
52                 dp[i][j] = max(dp[i][j],dp[i][k]+dp[k+1][j]+pre[j]-pre[i]+a[i]);
53                 dp2[i][j] = min(dp2[i][j],dp2[i][k]+dp2[k+1][j]+pre[j]-pre[i]+a[i]);
54             }
55         }
56     }
57     
58     int rnt = 0;
59     _for(i,0,N)
60         rnt = max(rnt,dp[i][i+N-1]);
61         
62     int rnt2 = 0x3f3f3f3f;
63     _for(i,0,N)
64         rnt2 = min(rnt2,dp2[i][i+N-1]);
65     write(rnt2);printf("\n");
66     write(rnt);
67     return 0;
68 }

 

posted @ 2019-08-10 10:45  Asurudo  阅读(130)  评论(0)    收藏  举报