hdu 5386 Cover

Description

You have an 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 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

For each case:

First line has two integer ,

Then lines,every line has integers,describe the initial matrix

Then lines,every line has integers,describe the goal matrix

Then lines,every line describe an operating

Output

For each case,print a line include integers.The i-th integer x show that the rank of x-th operating is

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

题目大意：给你n×n的初始图形，和目标图形，还有m个操作，操作和一把一行或一列变成一种颜色。如今问使初始图形变成目标图形的操作的顺序。每一个操作都要用上，且一定有解。

解题思路：初始图形没实用。直接从目标图形開始进行操作。

对于一个操作。推断该操作相应的那一行或一列的颜色，是否除0之外所有都是与该操作变换的颜色同样，是的话，将那一行或一列所有变为0，而且记录该操作，最后逆序输出。

注意，把所有操作遍历一遍是不够的。由于有一些操作是要在别的操作已进行过的基础上才干进行。所以在遍历操作的外层要再加一层循环（cnt

#include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           using namespace std;const int N = 205;const int M = 2005;typedef long long ll;int ans[M];int G1[N][N], G2[N][N];struct Node{    int pos, val, id;}H[M], L[M];int n, m, cntH, cntL;void init() {    memset(L, 0, sizeof(L));    memset(H, 0, sizeof(H));    memset(G2, 0, sizeof(G2));    cntH = cntL = 0;}void input() {    scanf("%d %d", &n, &m);    int a, b;    for (int i = 1; i <= n; i++) {        for (int j = 1; j <= n; j++) {            scanf("%d", &G1[i][j]);        }       }    for (int i = 1; i <= n; i++) {        for (int j = 1; j <= n; j++) {            scanf("%d", &G2[i][j]);         }       }    char s[20];    for (int i = 1; i <= m; i++) {        scanf("%s", s);        scanf("%d %d", &a, &b);        if (strcmp(s, "H") == 0) {            H[cntH].pos = a;                H[cntH].val = b;            H[cntH++].id = i;        } else {            L[cntL].pos = a;            L[cntL].val = b;            L[cntL++].id = i;        }    }}void solve() {    int cnt = 0;    while (cnt < m) {        for (int i = 0; i < cntH; i++) {            int k = H[i].pos, flag = 1;            if (!k) continue;             for (int j = 1; j <= n; j++) {                if(G2[k][j] && G2[k][j] != H[i].val) {                    flag = 0;                     break;                  }            }            if(flag) {                  ans[cnt++] = H[i].id;                  for(int j = 1; j <= n; j++) G2[k][j] = 0;                  H[i].pos = 0;            }          }        for (int i = 0; i < cntL; i++) {            int k = L[i].pos, flag = 1;              if (!k) continue;            for(int j = 1; j <= n; j++) {                if(G2[j][k] && G2[j][k] != L[i].val) {                    flag = 0;                     break;                  }            }            if(flag) {                  ans[cnt++] = L[i].id;                  for(int j = 1; j <= n; j++) G2[j][k] = 0;                  L[i].pos = 0;            }          }    }    for (int i = m - 1; i >= 1; i--) printf("%d ", ans[i]);    printf("%d\n", ans[0]);}int main() {    int T;    scanf("%d", &T);    while (T--) {        init();        input();            solve();    }    return 0;}