SGU 168 Matrix （DP）

原题：

      

168. Matrix

time limit per test: 0.5 sec. 
memory limit per test: 16000 KB

input: standard 
output: standard

You are given N*M matrix A. You are to find such matrix B, that B[i,j]=min{ A[x,y] : (y>=j) and (x>=i+j-y) }

Input

On the first line of the input there are two integer numbers, N and M (1<=N,M<=1000). Then matrix A follows: next N lines contains M integers each (not greater than 32000 by absolute value). The j-th number on then i-th of this lines is A[i,j].

Output

Write matrix B in the same format as matrix A, but without N and M.

Sample test(s)

Input

3 3 1 2 3 4 5 6 7 8 9

Output

1 2 3 
2 3 6 
3 6 9

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Author:NNSU #2 teamResource:
Date:

#include <iostream>#include <iomanip>#include <algorithm>#include <cstdio>#include <cstring>#include <queue>#include <deque>#include <string>#include <cmath>#include <stack>#include <vector>#include <utility>#include <set>#include <map>#include <sstream>#include <climits>#pragma comment(linker, "/STACK:1024000000,1024000000")#define pi acos(-1.0)#define INF 2147483647using namespace std;typedef long long ll;typedef pair <int,int > PP;int N[1005][1005],M[1005][1005],s[1005][1005];int main(){    int n,m;    scanf("%d%d",&n,&m);    int min_=32005;    for(int i=0; i<n; i++)        for(int j=0; j<m; j++)            scanf("%d",&N[i][j]);    for(int j=m-1; j>=0; j--)    {        int min__=32005;        for(int i=n-1; i>=0; i--)        {            min_=min(min_,N[i][j]);            min__=min(min__,N[i][j]);            s[i][j]=min__;        }        M[0][j]=min_;    }    for(int i=1; i<n; i++)        for(int j=0; j<m; j++)        {            min_=s[i][j];            if(j<m-1)            {                min_=min(min_,M[i-1][j+1]);            }            M[i][j]=min_;        }    for(int i=0;i<n;i++)    {        for(int j=0;j<m;j++)        {            j==0?printf("%d",M[i][j]):printf(" %d",M[i][j]);        }        printf("\n");    }}

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