hdu5386 Cover

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

这题看了题解后，感觉挺水的。。因为保证有解，所以可以从后面往前推，遇到整行的颜色和其中没有访问过的一个操作一样的时候，就把这一行的数都变为0（即任意颜色，因为前面的颜色会被后面的覆盖），当矩阵全部为0就输出结果。这里如果用set存储的话注意操作符的定义，因为如果定义为x或者y间的比较，可能会把一些相同的操作删除掉，导致WA.

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
int gra[106][106],c[600];
struct node{
	int f,x,y,idx;
}b,temp;
bool operator <(node a,node b){
	return a.idx<b.idx;
}

set<node>myset;
set<node>::iterator it;

int main()
{
	int n,m,i,j,T,sum,a,x,y,tot,flag,f,t,flag1,idx,num1;
	char s[10];
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d",&n,&m);
		sum=n*n;
		for(i=1;i<=n;i++){
			for(j=1;j<=n;j++){
				scanf("%d",&a);
			}
		}
		for(i=1;i<=n;i++){
			for(j=1;j<=n;j++){
				scanf("%d",&gra[i][j]);
			}
		}
		myset.clear();
		for(i=1;i<=m;i++){
			scanf("%s%d%d",s,&x,&y);
            if(s[0]=='L'){
            	b.f=1;
            }
            else b.f=0;
            b.x=x;b.y=y;b.idx=i;
            myset.insert(b);
		}
		t=0;
		while(1)
		{
			//if(myset.size()==0)break;// || sum==0
			if(sum==0)break;
			flag=0;
			for(it=myset.begin();it!=myset.end();it++){
				temp=*it;
				x=temp.x;y=temp.y;f=temp.f;idx=temp.idx;
				if(f==1){
					flag1=1;tot=0;
					for(i=1;i<=n;i++){
						if(gra[i][x]==0)continue;
						if(gra[i][x]==y)tot++;
						else{
							flag1=0;break;
						}
					}
					if(tot==0 || flag1==0)continue;
					
					flag=1;
					for(i=1;i<=n;i++){
						if(gra[i][x]==0)continue;
						else {gra[i][x]=0;sum--;}
					}
					t++;c[t]=idx;
					myset.erase(it);break;
				}
				
				else if(f==0){
					flag1=1;tot=0;
					for(i=1;i<=n;i++){
						if(gra[x][i]==0)continue;
						if(gra[x][i]==y)tot++;
						else{
							flag1=0;break;
						}
					}
					if(tot==0 || flag1==0)continue;
					
					flag=1;
					for(i=1;i<=n;i++){
						if(gra[x][i]==0)continue;
						else {gra[x][i]=0;sum--;}
					}
					t++;c[t]=idx;
					myset.erase(it);break;
				}
			}
			if(!flag)break;
		}
		for(i=1;i<=m;i++){
			flag=0;
			for(j=1;j<=t;j++){
				if(i==c[j]){
					flag=1;break;
				}
			}
			if(flag==0){
				printf("%d ",i);
			}
		}
		for(i=t;i>=1;i--){
			if(i==1)printf("%d\n",c[i]);
			else printf("%d ",c[i]);
		}
	}
	return 0;
}
/*
100
3 7
2 2 1
2 3 3
2 1 3
3 2 2
1 1 2
1 1 1
L 2 3
L 1 3
H 2 1
H 3 3
L 4 3
L 3 2
H 3 1
*/