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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 966 Accepted Submission(s): 320
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

題意：給出兩個n*n的矩陣。一個作為初始矩陣，一個作為目標矩陣。給出m個操作，操作有兩種，

一種是“L。x。y”，代表我們要把x這一行賦成y，還有一種是“H，x，y”，代表要把x這一列賦成y，

問我們怎樣安排這些操作才幹把初始矩陣轉化成目標矩陣。

輸出方案，special judge

題解：最後一個操作肯定是把某一行或者某一列變成x,我們倒過來模擬，每次把最後一個操作找出來。即每次找到某一行

或者某一列不為0的數都同樣的，再找符合操作的。

#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<iostream>
#define N 110

using namespace std;
int a[N][N];

struct Cao {
    char s[2];
    int x,v;
    bool used;
} b[N*5];
int ans[N*5];
int n,m;

bool is_H(int i,int k) {
    int x=-1;
    int j=1;
    for(; j<=n; j++) {
        if(a[i][j]) {
            x=a[i][j];
            break;
        }
    }
    if(x==-1)return true;
    if(x!=k)return false;
    for(; j<=n; j++) {
        if(a[i][j]&&a[i][j]!=x)return false;
    }
    return true;
}

bool is_L(int i,int k) {
    int x=-1;
    int j=1;
    for(; j<=n; j++) {
        if(a[j][i]) {
            x=a[j][i];
            break;
        }
    }
    if(x==-1)return true;
    if(x!=k)return false;
    for(; j<=n; j++) {
        if(a[j][i]&&a[j][i]!=x)return false;
    }
    return true;
}

int main() {
    // freopen("test.in","r",stdin);
    int t;
    cin>>t;
    while(t--) {
        scanf("%d%d",&n,&m);
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
                scanf("%d",&a[i][j]);
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
                scanf("%d",&a[i][j]);
        for(int i=1; i<=m; i++) {
            scanf("%s%d%d",b[i].s,&b[i].x,&b[i].v);
            b[i].used=0;
        }
        for(int h=m; h>=1; h--) {
            for(int i=1; i<=m; i++) {
                if(b[i].used)continue;
                if(b[i].s[0]=='H'&&is_H(b[i].x,b[i].v)) {
                    ans[h]=i;
                    b[i].used=1;
                    int p=b[i].x;
                    for(int k=1; k<=n; k++)
                        a[p][k]=0;
                    break;
                } else if(b[i].s[0]=='L'&&is_L(b[i].x,b[i].v)) {
                    ans[h]=i;
                    b[i].used=1;
                    int p=b[i].x;
                    for(int k=1; k<=n; k++)
                        a[k][p]=0;
                    break;
                }
            }
        }
        for(int i=1; i<m; i++)
            printf("%d ",ans[i]);
        printf("%d\n",ans[m]);
    }
    return 0;
}

HDU 5386 Cover（模擬）