【鏈接】 我是鏈接,點我呀:)
【題意】

在這裏輸入題意

【題解】

先預處理出來一個正方形。
然後每次枚舉新加的正方形左上角的坐標就可以。
註意覆蓋的規則,控制一下就可以。
然後暴力判斷是否相同。
暴力回溯即可(只用回溯一個正方形區域)

【代碼】

/*
    1.Shoud it use long long ?
    2.Have you ever test several sample(at least therr) yourself?
    3.Can you promise that the solution is right? At least,the main ideal
 

    4.use the puts("") or putchar() or printf and such things?
    5.init the used array or any value?
    6.use error MAX_VALUE?
    7.use scanf instead of cin/cout?
    8.whatch out the detail input require
*/
/*
    一定在這裏寫完思路再敲代碼!!!
*/
#include <bits/stdc++.h>
using namespace std;

const
 
 int N = 10;
const int M = 5;

char s[N+5][N+5],now[N+5][N+5];
char square[M+5][M+5];

void init(){
    for (int i = 0;i <3;i++)
        for (int j = 0;j<5;j++)
            square[i][j] = ' ';
    for (int i = 0;i < 2;i++) square[i+1][0] = square[i+1][4] = '|';
    for (int j = 1
 
;j < 5;j+=2) square[0][j]=square[2][j] = '_';
}

void Set(int x,int y){
    for (int i = 0;i < 3;i++)
        for (int j = 0;j < 5;j++){
            if (i<=0 && now[i+x][j+y]!=' ' && square[i][j]==' ') continue;
            now[i+x][j+y] = square[i][j];
        }
    for (int i = 1;i < 2;i++)
        for (int j = 1;j<4;j++)
            now[i+x][j+y] = ' ';

}

bool ok(){
    for (int i = 0;i < 5;i++)
        for (int j = 0;j < 9;j++)
            if (now[i][j]!=s[i][j])
                return false;
    return true;
}

void out(){
    for (int i = 0;i < 5;i++){
        for (int j = 0;j < 9;j++)
            cout <<now[i][j];
        cout << endl;
    }
    cout << endl;

}

bool dfs(int dep){
    if (dep > 1 && ok()) return true;
    if (dep >= 7) return false;
    int temp[M+5][M+5];
    for (int i = 0;i < 3;i++){
        for (int j = 0;j < 5;j+=2){
            for (int k = 0;k <3;k++)
                for (int l = 0;l < 5;l++)
                    temp[k][l] = now[k+i][l+j];
            Set(i,j);
            if (dfs(dep+1)) return true;
            for (int k = 0;k <3;k++)
                for (int l = 0;l < 5;l++)
                    now[k+i][l+j] = temp[k][l];
        }
    }
    return false;
}

int main(){
    #ifdef LOCAL_DEFINE
        freopen("rush_in.txt", "r", stdin);
    #endif
    ios::sync_with_stdio(0),cin.tie(0);
    init();
    int kase = 0;
    while (1){
        for (int i = 0;i < 5;i++){
            cin.getline(s[i],15);
            if (s[i][0]=='0') return 0;
        }
        for (int i = 0;i< 5;i++)
            for (int j = 0;j < 9;j++)
                now[i][j] = ' ';
        if (dfs(1)){
            cout <<"Case "<<++kase<<": Yes"<<endl;
        }else{
            cout <<"Case "<<++kase<<": No"<<endl;
        }
    }
    return 0;
}

【習題 7-6 UVA - 12113】Overlapping Squares