【51nod】1291 Farmer 【求n*m的矩形内全1的i*j的矩形个数】【dp】

题目链接：【51nod】1291 Farmer

去年做的题，一直没有放出来，单调栈，O(N^2)。O(N^2logN)也能过。

#include <bits/stdc++.h>
using namespace std ;

#define clr( a , x ) memset ( a , x , sizeof a )

const int MAXN = 1005 ;

char s[MAXN] ;
int U[MAXN] , S[MAXN] , cur , c[MAXN][MAXN] , n , m ;

void solve () {
    clr ( U , 0 ) ;
    clr ( c , 0 ) ;
    for ( int i = 1 ; i <= n ; ++ i ) {
        scanf ( "%s" , s + 1 ) ;
        for ( int j = 1 ; j <= m ; ++ j ) U[j] = s[j] == '1' ? U[j] + 1 : 0 ;
        cur = 0 ;
        S[++ cur] = 0 ;
        for ( int j = 1 ; j <= m + 1 ; ++ j ) {
            while ( U[j] < U[S[cur]] ) {
                c[max ( U[S[cur - 1]] , U[j] ) + 1][j - S[cur - 1] - 1] ++ ;
                c[U[S[cur]] + 1][j - S[cur - 1] - 1] -- ;
                -- cur ;
            }
            while ( cur >= 1 && U[j] == U[S[cur]] ) -- cur ;
            S[++ cur] = j ;
        }
    }
    for ( int i = 1 ; i <= n ; ++ i ) {
        for ( int j = 1 ; j <= m ; ++ j ) c[i][j] += c[i - 1][j] ;
    }
    for ( int i = 1 ; i <= n ; ++ i ) {
        for ( int j = m - 1 ; j >= 1 ; -- j ) c[i][j] += c[i][j + 1] ;
        for ( int j = m - 1 ; j >= 1 ; -- j ) c[i][j] += c[i][j + 1] ;
    }
    for ( int i = 1 ; i <= n ; ++ i ) {
        for ( int j = 1 ; j <= m ; ++ j ) {
            printf ( "%d%c" , c[i][j] , j < m ? ' ' : '\n' ) ;
        }
    }
}

int main () {
    while ( ~scanf ( "%d%d" , &n , &m ) ) solve () ;
    return 0 ;
}