Codeforces 815 A Karen and Game 贪心

最新推荐文章于 2021-03-10 15:27:04 发布

weixin_33726318

最新推荐文章于 2021-03-10 15:27:04 发布

阅读量51

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/FriskyPuppy/p/7388346.html

本文解析了CodeForces上的一道题目，通过贪心算法求解一个n*m矩阵达到目标状态所需的最少操作次数及具体步骤。根据矩阵的维度大小不同，采取不同的策略优化计算过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

　　题目链接： http://codeforces.com/problemset/problem/815/A

　　题目描述：一个初始都为0的n * m 的格子，你做的操作只有两种，给一行+1，给一列加一，问给出你最终状态，输出最少步数和步骤

　　解题思路：简单贪心即可，自己一开始没有讨论n, m大小关系， WA了一发

　　代码：

#include <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iterator>
#include <cmath>
#include <algorithm>
#include <stack>
#include <deque>
#include <map>
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define mem0(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,0x3f,sizeof(a))
typedef long long ll;
using namespace std;

const int INF = 0x3fffffff;
const int maxn = 510;
int g[maxn][maxn];
int row[maxn];
int t1[maxn];
int col[maxn];
int t2[maxn];

int main() {
    int n, m;
    int sum = 0;
    mem0(g);
    mem0(row);
    mem0(col);
    mem0(t1);
    mem0(t2);
    scanf( "%d%d", &n, &m );
    for( int i = 1; i <= n; i++ ) {
        for( int j = 1; j <= m; j++ ) {
            scanf( "%d", &g[i][j] );
        }
    }
    int cnt = 0;
    int cnt2 = 0;
    if( n > m ) {
        for( int j = 1; j <= m; j++ ) {
            int minum = INF;
            for( int i = 1; i <= n; i++ ) {
                minum = min( minum, g[i][j] );
            }
            sum += minum;
            if( minum > 0 ) {
                col[cnt2] = j;
                t2[cnt2++] = minum;
            }
            for( int i = 1; i <= n; i++ ) {
                g[i][j] -= minum;
            }
        }
        for( int i = 1; i <= n; i++ ) {
            int minum = INF;
            for( int j = 1; j <= m; j++ ) {
                minum = min( minum, g[i][j] );
            }
            sum += minum;
            if( minum > 0 ) {
                row[cnt] = i;
                t1[cnt++] = minum;
            }
            for( int j = 1; j <= m; j++ ) {
                g[i][j] -= minum;
            }
        }
    }
    else {
        for( int i = 1; i <= n; i++ ) {
            int minum = INF;
            for( int j = 1; j <= m; j++ ) {
                minum = min( minum, g[i][j] );
            }
            sum += minum;
            if( minum > 0 ) {
                row[cnt] = i;
                t1[cnt++] = minum;
            }
            for( int j = 1; j <= m; j++ ) {
                g[i][j] -= minum;
            }
        }
        for( int j = 1; j <= m; j++ ) {
            int minum = INF;
            for( int i = 1; i <= n; i++ ) {
                minum = min( minum, g[i][j] );
            }
            sum += minum;
            if( minum > 0 ) {
                col[cnt2] = j;
                t2[cnt2++] = minum;
            }
            for( int i = 1; i <= n; i++ ) {
                g[i][j] -= minum;
            }
        }
    }
    for( int i = 1; i <= n; i++ ) {
        for( int j = 1; j <= m; j++ ) {
            if( g[i][j] != 0 ) {
                printf( "-1\n" );
                return 0;
            }
        }
    }
    printf( "%d\n", sum );
    for( int i = 0; i < cnt; i++ ) {
        while( t1[i]-- ) {
            printf( "row %d\n", row[i] );
        }
    }
    for( int i = 0; i < cnt2; i++ ) {
        while( t2[i]-- ) {
            printf( "col %d\n", col[i] );
        }
    }
    return 0;
}