12113:Overlapping Squares

Overlapping Squares

我的思路：可以根据所给图形计算出图形中包含的方块的个数以及它们各自的位置，方块可以根据某一个角点的位置来确定，具体方法见 count() 函数。方块确定之后，图形的各种变化就取决于方块们的不同放置顺序了，枚举所有排列，进行模拟，看看模拟结果里面有没有所给图形即可。

思路没什么问题，代码了检查了好多遍，但是一直WA，暂时还没发现问题，再想想吧。

#include
using namespace std;
const int maxn = 10;
const int m = 5, n = 9;
typedef pair P;
P ps[maxn];
int ok = 0;
int ns, vis[maxn], found[m][n];
char G[m][n+5], R[m][n+1];

int count(){
    int cnt = 0;
    for(int i = 0; i < m; i++){
        for(int j = 0; j < n; j++){
            if(i < m-2 && j < n-4 && G[i][j+1] == '_' && G[i+1][j] == '|'){
                found[i][j] =  1;
                ps[cnt++] = P(i, j);
            }
            else if(i < m-2 && j >= 4 && G[i][j-1] == '_' && G[i+1][j] == '|'){
                if(found[i][j-4]) continue;
                found[i][j-4] = 1;
                ps[cnt++] = P(i, j - 4);
            }
            else if(i >= 2 && j < n-4 && G[i][j+1] == '_' && G[i-1][j] == '|'){
                if(found[i-2][j]) continue;
                found[i-2][j] = 1;
                ps[cnt++] = P(i - 2, j);
            }
            else if(i >= 2 && j >= 4 && G[i][j-1] == '_' && G[i-1][j] == '|'){
                if(found[i-2][j-4]) continue;
                found[i-2][j-4] = 1;
                ps[cnt++] = P(i - 2, j - 4);
            }
        }
    }
    return cnt;
}

int DFS(int i, int num){
    int x = ps[i].first, y = ps[i].second;
    for(int i = 1; i <= 2; i++){
        R[x+i][y] = R[x+i][y+4] = '|';
    }
    for(int i = 1; i <= 3; i += 2){
        R[x][y+i] = R[x+2][y+i] = '_';
    }
    for(int j = 1; j <= 3; j++){
        R[x+1][y+j] = ' ';
        if(R[x+2][y+j] == '|') R[x+2][y+j] = ' ';
    }
    if(num == ns){
        int ok = 1;
        for(int i = 0; i < m && ok; i++){
            for(int j = 0; j < n && ok; j++){
                if(G[i][j] != R[i][j]) ok = 0;
            }
        }
        return ok;
    }

    for(int i = 0; i < ns; i++){
        if(!vis[i]){
            vis[i] = 1;
            if(DFS(i, num + 1)) return 1;
            vis[i] = 0;
        }
    }
    return 0;
}

int main()
{
    freopen("data.in","r",stdin);
    freopen("data.out","w",stdout);
    int T = 0;
    while(fgets(G[0], 15, stdin) && G[0][0] != '0'){
        for(int i = 1; i < m; i++){
            fgets(G[i], 15, stdin);
        }
        memset(found, 0, sizeof(found));
        ns = count();
        // printf("\nns: %d\n", ns);
        ok = 0;
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                R[i][j] = ' ';
            }
            R[i][n] = 0;
        }
        if(ns >= 1 && ns <= 6){
            for(int i = 0; i < ns; i++){
                vis[i] = 1;
                if(DFS(i, 1)){ ok = 1; break; }
                vis[i] = 0;
            }
        }
        printf("Case %d: %s\n", ++T, ok ? "Yes" : "No");
    }
    return 0;
}