http://poj.org/problem?id=1050
最大連續子序列的二維版本,枚舉第i、j行,之間的元素和構成一個n個元素的序列,求其最大連續子序列即可,維護maxx,複雜度O(n^3)
/*
poj 1050 最大子矩陣和dp
(一維)最大連續子序列和的拓展(二維)
思路:
暴力枚舉所有子矩陣求和找最大值,O(n^4)超時
將二維降成一維,枚舉第i,j行,i-j行每列的元素和作爲一個元素,求其構成的最大連續子序列和
整個過程中維護一個maxx即可
拿前綴和預處理了下,複雜度O(n^3)10^6 16ms飄過
*/
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<vector>
#include<queue>
#include<set>
#include<map>
#include<algorithm>
#include<sstream>
#define eps 1e-9
#define pi acos(-1)
#define INF 0x7fffffff
#define inf -INF
#define MM 12900
#define N 50
using namespace std;
typedef long long ll;
const int _max = 100 + 10;
int a[_max][_max],n,col[_max][_max],b[_max];
int dp[_max];
int work(){//最大連續子序列和O(n)
dp[1] = b[1];//dp[i]以i爲結尾的最大連續子序列和
int maxx = dp[1];
for(int i = 2; i <= n; ++ i) {
if(dp[i-1]+b[i]<b[i]) dp[i] = b[i];
else dp[i] = dp[i-1] + b[i];
if(maxx < dp[i]) maxx = dp[i];
}
return maxx;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("input.txt","r",stdin);
#endif // ONLINE_JUDGE
while(scanf("%d",&n)==1){
for(int i = 1; i <= n; ++ i)
for(int j = 1; j <= n; ++ j)
scanf("%d",&a[i][j]);
//前綴和預處理
memset(col,0,sizeof(col));
for(int i = 1; i <= n;++ i) col[i][1] = a[1][i];
for(int i = 1; i <= n; ++ i)//列
for(int j = 2; j <= n; ++ j)
col[i][j] = col[i][j-1] + a[j][i];
//O(n^3)
int t,maxx = inf;//會枚舉到所有子矩陣,inf即可,不用a[1][1]
for(int i = 1;i <= n; ++ i){//枚舉i-j行
for(int j = i; j <= n; ++ j){
b[1] = col[1][j]-col[1][i-1];
for(int k = 2; k <= n; ++ k)
b[k] = col[k][j] - col[k][i-1];
t = work();
if(maxx<t) maxx = t;
}
}
printf("%d\n",maxx);
}
return 0;
}