标签:turn sizeof its com onclick clu sed space isp
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12471 Accepted Submission(s): 5985
#include <iostream> #include <algorithm> #include <vector> #include<string.h> using namespace std; int a[55][55]; int main() { int n,m; cin>>n>>m; for(int i=0;i<n;i++) { for(int j=0;j<m;j++) { cin>>a[i][j]; a[i][j]+=a[i-1][j]; } } int dp[55]; memset(dp,0,sizeof(dp)); int ans=-0x3fffffff; for(int i=0;i<n-1;i++) { for(int j=i+1;j<n;j++) { for(int k=1;k<=m;k++) { dp[k]=a[j][k-1]-a[i][k-1]; dp[k]=dp[k]+dp[k-1]; if(dp[k]>ans) ans=dp[k]; if(dp[k]<0) dp[k]=0; } } } printf("%d\n",ans); return 0; }
令a[i][k]保存第k列,前i行的和。这样将矩阵通过a[j][k]-a[i][k]压缩成一维,然后求最大字串和。。
标签:turn sizeof its com onclick clu sed space isp
原文地址:http://www.cnblogs.com/superxuezhazha/p/6387756.html