【codeforces 816C】Karen and Game
2017-10-04 18:44
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【题目链接】:http://codeforces.com/contest/816/problem/C
【题意】
给你一个n*m的矩阵;
一开始所有数字都是0;
每次操作,你能把某一行,或某一列的数字全部加上1;
问你到达目标矩阵最少需要进行多少次操作;
【题解】
从目标矩阵开始减;
直接枚举每一列需要减多少次;
每一行需要减多少次即可;
但有技巧;
比如以下矩阵
1 1 1
1 1 1
1 1 1
1 1 1
应该一列一列地减比较快;
而
1 1 1 1
1 1 1 1
一行一行地删比较快;
所以对n和m的大小关系分类讨论一下;
看是先删行还是先删列;
【Number Of WA】
4
【反思】
一开始不知道在想什么,光想着快点写完了,结果写了个没任何把握的贪心,一直瞎试。
没想到多试几个输入hack一下自己;太着急了;
列的英文是colum/xk
【完整代码】
#include <bits/stdc++.h> using namespace std; #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define LL long long #define rep1(i,a,b) for (int i = a;i <= b;i++) #define rep2(i,a,b) for (int i = a;i >= b;i--) #define mp make_pair #define pb push_back #define fi first #define se second #define ms(x,y) memset(x,y,sizeof x) #define Open() freopen("F:\\rush.txt","r",stdin) #define Close() ios::sync_with_stdio(0) typedef pair<int,int> pii; typedef pair<LL,LL> pll; const int dx[9] = {0,1,-1,0,0,-1,-1,1,1}; const int dy[9] = {0,0,0,-1,1,-1,1,-1,1}; const double pi = acos(-1.0); const int N = 110; int n,m,ans = 0; int g ,srow ,sclu ; vector <int> row,col; void subrow(){ rep1(i,1,n){ int mi = 1000; rep1(j,1,m){ mi = min(mi,g[i][j]); } rep1(j,1,m) g[i][j]-=mi; srow[i] = mi; ans+=mi; } } void subclu(){ rep1(j,1,m){ int mi = 1000; rep1(i,1,n){ mi = min(mi,g[i][j]); } ans+=mi; rep1(i,1,n) g[i][j]-=mi; sclu[j] = mi; } } int main(){ //Open(); Close(); cin >> n >> m; rep1(i,1,n){ rep1(j,1,m) cin >> g[i][j]; } if (n < m){ subrow(); subclu(); }else { subclu(); subrow(); } rep1(i,1,n) rep1(j,1,m) if (g[i][j]) return cout <<-1<<endl,0; cout << ans << endl; rep1(i,1,n){ while (srow[i]--){ cout <<"row "<<i<<endl; } } rep1(j,1,m){ while (sclu[j]--){ cout <<"col "<<j<<endl; } } return 0; }
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