[貪心] hdu5386 多校聯合第八場 Cover

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Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1049    Accepted Submission(s): 217
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

我們只要每次找一行或一列顏色除了$0$都相同的，然後如果有對應的操作，就把這行這列都賦值成$0$即可

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<set>
#include<vector>
#include<cmath>
using namespace std;

int INF =0x3f3f3f3f;
int cur[105][105], goal[105][105];
bool vis[105][105];
int n, m;
struct nde
{
    char c;
    int x, y, id;
    int sortlist;
} node[555];

bool cmp(nde a, nde b)
{
    return a.sortlist<b.sortlist;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        INF = 0x3f3f3f3f;
        memset(node,0,sizeof node);
        memset(vis,0,sizeof vis);

        scanf("%d%d",&n,&m);
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
                scanf("%d",&cur[i][j]);

        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
                scanf("%d",&goal[i][j]);

        for(int i=0; i<m; i++)
        {
            char s[10];
            scanf("%s%d%d",s,&node[i].x,&node[i].y);
            node[i].c=s[0];
            node[i].id=i+1;
        }
        int temp = m;
        while(temp--)
        {
            for(int i=0; i<m; i++)
            {
                if(node[i].sortlist>0) continue;
                bool flag=true;
                if(node[i].c=='L')
                {
                    //如果修改過了該值或者該值就是原來的 那就不用再改了
                    for(int j=1; j<=n; j++)
                    if(!(goal[j][node[i].x] == node[i].y || vis[j][node[i].x]))
                        flag=false;
                    if(flag)
                    {
                        for(int j=1; j<=n; j++)
                            vis[j][node[i].x]=true; //訪問過了
                        node[i].sortlist=temp;
                        break;
                    }
                }
                else if (node[i].c =='H')
                {
                    for(int j=1; j<=n; j++)
                    {
                    if(!(goal[node[i].x][j] == node[i].y || vis[node[i].x][j]))
                        flag=false;
                    }
                    if(flag)
                    {
                        for(int j=1; j<=n; j++)
                            vis[node[i].x][j]=true;
                        node[i].sortlist=temp;
                        break;
                    }
                }
            }
        }

        sort(node,node+m,cmp);

        int i = 0;
        for(i=0; i<m-1; i++)
            printf("%d ",node[i].id);
        printf("%d\n",node[i].id);
    }
    return 0;
}