BZOJ 1084 最大子矩阵

http://www.lydsy.com/JudgeOnline/problem.php?id=1084

思路:分m=1和m=2操作

 1 #include<algorithm>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<iostream>
 6 int f[1005][1005],F[1005][1005][11];
 7 int sum[10005],sum1[10005],sum2[10005],n,m,K;
 8 int read(){
 9     int t=0,f=1;char ch=getchar();
10     while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
11     while ('0'<=ch&&ch<='9'){t=t*10+ch-'0';ch=getchar();}
12     return t*f;
13 }
14 void sxpianfen1(){
15     for (int i=1;i<=n;i++)
16      {int x=read();sum[i]=sum[i-1]+x;}      
17     for (int i=1;i<=n;i++)
18      for (int k=1;k<=K;k++){
19       f[i][k]=f[i-1][k];
20       for (int j=0;j<i;j++)
21        f[i][k]=std::max(f[i][k],f[j][k-1]+sum[i]-sum[j]);
22     } 
23     printf("%d\n",f[n][K]);
24 }
25 void sxpianfen2(){
26     for (int i=1;i<=n;i++){
27         int x=read(),y=read();
28         sum1[i]=sum1[i-1]+x;
29         sum2[i]=sum2[i-1]+y;
30     }
31     for (int k=1;k<=K;k++)
32      for (int i=1;i<=n;i++)
33       for (int j=1;j<=n;j++){
34          F[i][j][k]=std::max(F[i-1][j][k],F[i][j-1][k]);
35          for (int l=0;l<i;l++) F[i][j][k]=std::max(F[i][j][k],F[l][j][k-1]+sum1[i]-sum1[l]);
36          for (int l=0;l<j;l++) F[i][j][k]=std::max(F[i][j][k],F[i][l][k-1]+sum2[j]-sum2[l]);
37          if (i==j)
38          for (int l=0;l<i;l++) F[i][j][k]=std::max(F[i][j][k],F[l][l][k-1]+sum1[i]+sum2[j]-sum1[l]-sum2[l]);
39       }
40     printf("%d\n",F[n][n][K]);  
41 }
42 int main(){
43     n=read();m=read();K=read();
44     if (m==1){
45         sxpianfen1();
46     }else
47         sxpianfen2();
48 }

 

posted @ 2016-06-16 20:56  GFY  阅读(58)  评论(0编辑  收藏