算典04_习题_02_UVA-201

正方形

题意

    给你一个正方形，每行和每列都有n个点，有些点是相连的，有些点是不相连的，一个点只会跟它上下左右四个点中的一些点相连。让你统计总共有多少个正方形，并输出各种规模的正方形各有多少个

题解

    由于题目规模较小，直接暴力枚举即可，即分别枚举每种规模的正方形出现的次数

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define met(a, b) memset(a, b, sizeof(a));
const int maxn = 1e1 + 5;

int n, m, a, b;
bool maps[maxn][maxn][maxn][maxn];
char c;

bool judge(int x, int y, int len) {
    for(int i = 0; i < len; ++i) if(!maps[x][y+i][x][y+i+1])return 0;
    for(int i = 0; i < len; ++i) if(!maps[x+len][y+i][x+len][y+i+1])return 0;
    for(int i = 0; i < len; ++i) if(!maps[x+i][y][x+i+1][y])return 0;
    for(int i = 0; i < len; ++i) if(!maps[x+i][y+len][x+i+1][y+len])return 0;
    return 1;
}

int solve(int len) {
    int res = 0;
    for(int i = 1; i <= n-len; ++i) {
        for(int j = 1; j <= n-len; ++j) {
            if(judge(i, j, len)) ++res;
        }
    }
    return res;
}

int main() {
    #ifdef _LOCAL
    freopen("in.txt", "r", stdin);
    #endif // _LOCAL

    int kase = 0, line = 0;
    while(scanf("%d", &n) == 1) {
        met(maps, 0);
        scanf("%d", &m);
        for(int i = 0; i < m; ++i) {
            cin >> c >> a >> b;
            if(c == 'H') maps[a][b][a][b+1] = 1;
            else maps[b][a][b+1][a] = 1;
        }

        if(line) printf("\n**********************************\n\n"); if(!line) line = 1;
        printf("Problem #%d\n\n", ++kase);
        int ans = 0, ok = 1;
        for(int i = 1; i <= n; ++i) {
            if(ans = solve(i)) printf("%d square (s) of size %d\n", ans, i), ok = 0;
        }
        if(ok) printf("No completed squares can be found.\n");
    }
    return 0;
}