hdu5386 Cover（逆向思维+模拟）

Link：http://acm.hdu.edu.cn/showproblem.php?pid=5386

Cover

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1247    Accepted Submission(s): 430
Special Judge

Problem Description

You have an  n∗n  matrix.Every grid has a color.Now there are two types of operating:
L x y: for(int i=1;i<=n;i++)color[i][x]=y;
H x y:for(int i=1;i<=n;i++)color[x][i]=y;
Now give you the initial matrix and the goal matrix.There are  m  operatings.Put in order to arrange operatings,so that the initial matrix will be the goal matrix after doing these operatings

It's guaranteed that there exists solution.

 

Input

There are multiple test cases,first line has an integer  T
For each case:
First line has two integer  n , m
Then  n  lines,every line has  n  integers,describe the initial matrix
Then  n  lines,every line has  n  integers,describe the goal matrix
Then  m  lines,every line describe an operating

1≤color[i][j]≤n
T=5
1≤n≤100
1≤m≤500

 

Output

For each case,print a line include  m  integers.The i-th integer x show that the rank of x-th operating is  i

 

Sample Input

  

   1
3 5
2 2 1 
2 3 3 
2 1 3 
3 3 3 
3 3 3 
3 3 3 
H 2 3
L 2 2
H 3 3
H 1 3
L 2 3
  

 

Sample Output

  

   5 2 4 3 1
  

 

Author

SXYZ

 

Source

2015 Multi-University Training Contest 8

 

官方题解：

1007.Cover

水题，我们只要每次找一行或一列颜色除了$0$都相同的，然后如果有对应的操作，就把这行这列都赋值成$0$即可

其实初始矩阵并没有什么卵用

稍微优化下复杂度是O(n^2)O(n​2​​)的，但是为了简易些范围开到了$100$

AC code：

#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
#define LL long long
#define MAXN 1000010 
using namespace std;
struct node{
	char ch[3];
	int x;
	int y;
}op[MAXN];
int color1[111][111];
int color2[111][111];
bool vis[MAXN];
int ans[MAXN];
int main()
{
	//freopen("D:\in.txt","r",stdin);
	int n,m,t,i,j,k,cnt;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=n;j++)
			{
				scanf("%d",&color1[i][j]);
			}
		}
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=n;j++)
			{
				scanf("%d",&color2[i][j]);
			}
		}
		for(i=1;i<=m;i++)
		{
			scanf("%s%d%d",op[i].ch,&op[i].x,&op[i].y);
		}
	/*	for(i=1;i<=m;i++)
		{
			printf("%s %d %d\n",op[i].ch,op[i].x,op[i].y);
		}*/
		memset(vis,false,sizeof(vis));
		cnt=0;
		while(1)
		{
			for(i=1;i<=m;i++)
			{
				if(vis[i])
					continue;
				if(op[i].ch[0]=='H')
				{
					j=op[i].x;
					for(k=1;k<=n;k++)
					{
						if(color2[j][k]!=0&&color2[j][k]!=op[i].y)
							break;
					}
					if(k==n+1)
					{
						for(k=1;k<=n;k++)
						{
							color2[j][k]=0;
						}
						ans[++cnt]=i;
						vis[i]=true;
					}
				}
				else if(op[i].ch[0]=='L')
				{
					j=op[i].x;
					for(k=1;k<=n;k++)
					{
						if(color2[k][j]!=0&&color2[k][j]!=op[i].y)
							break;
					}
					if(k==n+1)
					{
						for(k=1;k<=n;k++)
						{
							color2[k][j]=0;
						}
						ans[++cnt]=i;
						vis[i]=true;
					}
				}
				if(cnt==m)
					break;
			}
			if(cnt==m)
				break;
		}
		for(i=cnt;i>=1;i--)//注意：因为顺序是倒着推出来的，所以输出时也要逆着输出 
		{
			if(i==cnt)
				printf("%d",ans[i]);
			else
				printf(" %d",ans[i]);
		}
		puts("");
	}
	return 0;
 }