hdu 5386 模拟

想明白以后会发现其实就是模拟...

因为每次只能给一整行或者一整列赋值，所以目标矩阵一开始一定有一行或者一列上数字都是相同的，找到对应的操作，把那一行或者那一列标记为访问过（即从原矩阵中删除）。然后新的矩阵也一定能找到一行或者一列数字都相同，再找到相应的操作，标记那一行或者那一列，依次类推，然后没有用到的操作随便找个顺序就好了。其实初始矩阵并没有什么用......

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int N = 101;
const int M = 501;
int mp1[N][N];
int mp2[N][N];
int goal[N][N];
bool visit[N][N];
bool used[M];
int ans[M];
int n, m, cnt;

struct Node
{
    char op[2];
    int x, y;
} node[M];

int checkRow( int row )
{
    int j = 1;
    while ( j <= n && visit[row][j] ) j++;
    if ( j == n + 1 ) return -1;
    int num = goal[row][j];
    for ( int k = j + 1; k <= n; k++ )
    {
        if ( visit[row][k] ) continue;
        if ( num != goal[row][k] ) return -1;
    }
    return num;
}

int checkCol( int col )
{
    int i = 1;
    while ( i <= n && visit[i][col] ) i++;
    if ( i == n + 1 ) return -1;
    int num = goal[i][col];
    for ( int k = i + 1; k <= n; k++ )
    {
        if ( visit[k][col] ) continue;
        if ( num != goal[k][col] ) return -1;
    }
    return num;
}

void solve()
{
    memset( visit, 0, sizeof(visit) );
    memset( used, 0, sizeof(used) );
    cnt = 1;
    while ( 1 )
    {
        bool flag = false;
        for ( int row = 1; row <= n; row++ )
        {
            int tmp = checkRow(row);
            if ( tmp != -1 && mp1[row][tmp] != -1 )
            {
                ans[cnt++] = mp1[row][tmp];
                used[mp1[row][tmp]] = true;
                for ( int k = 1; k <= n; k++ )
                {
                    visit[row][k] = 1;
                }
                flag = true;
            }
        }
        for ( int col = 1; col <= n; col++ )
        {
            int tmp = checkCol(col);
            if ( tmp != -1 && mp2[col][tmp] != -1 )
            {
                ans[cnt++] = mp2[col][tmp];
                used[mp2[col][tmp]] = true;
                for ( int k = 1; k <= n; k++ )
                {
                    visit[k][col] = 1;
                }
                flag = true;
            }
        }
        if ( !flag ) break;
    }
    for ( int i = 1; i <= m; i++ )
    {
        if ( !used[i] )
        {
            ans[cnt++] = i;
        }
    }
    for ( int i = m; i > 1; i-- )
    {
        printf("%d ", ans[i]);
    }
    printf("%d
", ans[1]);
}

int main()
{
    int t;
    scanf("%d", &t);
    while ( t-- )
    {
        scanf("%d%d", &n, &m);
        for ( int i = 1; i <= n; i++ )
        {
            for ( int j = 1; j <= n; j++ )
            {
                scanf("%d", &goal[i][j]);
            }
        }
        for ( int i = 1; i <= n; i++ )
        {
            for ( int j = 1; j <= n; j++ )
            {
                scanf("%d", &goal[i][j]);
            }
        }
        memset( mp1, -1, sizeof(mp1) );
        memset( mp2, -1, sizeof(mp2) );
        for ( int i = 1; i <= m; i++ )
        {
            scanf("%s %d %d", &node[i].op, &node[i].x, &node[i].y);
            if ( node[i].op[0] == 'H' )
            {
                mp1[node[i].x][node[i].y] = i;
            }
            else
            {
                mp2[node[i].x][node[i].y] = i;
            }
        }
        solve();
    }
    return 0;
}