HDU 5386 Cover

Cover

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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

13 52 2 1 2 3 3 2 1 3 3 3 3 3 3 3 3 3 3 H 2 3L 2 2H 3 3H 1 3L 2 3

 

Sample Output

5 2 4 3 1

 

想复杂了，暴力解决即可，原始数组没有，从目标数组倒着思考每一个操作，多次重复，排序即可。

#pragma comment(linker, "/STACK:1024000000,1024000000")#include <set>#include <map>#include <stack>#include <cmath>#include <queue>#include <cstdio>#include <bitset>#include <string>#include <vector>#include <iomanip>#include <cstring>#include <iostream>#include <algorithm>#include <functional>#define inf 0x7fffffff#define maxn 100010using namespace std;struct OP{        char s;        int c, v;}op[550];int a[110][110];int b[110][110];int vis[550], ran[550];int main(){        int t;        scanf("%d", &t);        while (t--)        {                int n, m;                scanf("%d %d", &n, &m);                for (int i = 0; i < n; i++)                {                        for (int j = 0; j < n; j++)                        {                                scanf("%d", &a[i][j]);                        }                }                for (int i = 0; i < n; i++)                {                        for (int j = 0; j < n; j++)                        {                                scanf("%d", &b[i][j]);                        }                }                memset(vis, 0, sizeof(vis));                for (int i = 0; i < m; i++)                {                        scanf(" %c %d %d", &op[i].s, &op[i].c, &op[i].v);                }                int cnt = 1;                while (cnt <= m)                {                        for (int i = 0; i < m; i++)                        {                                if (vis[i])                                {                                        continue;                                }                                if (op[i].s == 'H')                                {                                        int flag = 0;                                        int pos = op[i].c-1;                                        for (int j = 0; j < n; j++)                                        {                                                if (b[pos][j] != 0 && b[pos][j] != op[i].v)                                                {                                                        flag = 1;                                                        break;                                                }                                        }                                        if (flag == 0)                                        {                                                vis[i] = 1;                                                ran[cnt++] = i;                                                for (int j = 0; j < n; j++)                                                {                                                        b[pos][j] = 0;                                                }                                        }                                }                                else                                {                                        int flag = 0;                                        int pos = op[i].c-1;                                        for (int j = 0; j < n; j++)                                        {                                                if (b[j][pos] != 0 && b[j][pos] != op[i].v)                                                {                                                        flag = 1;                                                        break;                                                }                                        }                                        if (flag == 0)                                        {                                                vis[i] = 1;                                                ran[cnt++] = i;                                                for (int j = 0; j < n; j++)                                                {                                                        b[j][pos] = 0;                                                }                                        }                                }                        }                }                for (int i = m; i >= 1; i--)                {                        if (i != 1)                        {                                printf("%d ", ran[i] + 1);                        }                        else                        {                                printf("%d\n", ran[i] + 1);                        }                }        }        return 0;}