习题7-16(uva-11846)

#include <iostream>
#include <istream>
#include <sstream>
#include <vector>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <cstring>
#include <unordered_map>
#include <unordered_set>
#include <algorithm>
#include <numeric>
#include <chrono>
#include <ctime>
#include <cmath>
#include <cctype>
#include <string>
#include <cstdio>
#include <iomanip>


#include <thread>
#include <mutex>
#include <condition_variable>
#include <functional>
#include <iterator>
using namespace std;
const int maxn = 25;
char matrix[maxn][maxn], ans[maxn][maxn];
int n, k;

int GetGroupCnt(int l, int t, int r, int d) {
	int cnt = 0,ret = 0;
	for (int i = l; i <= r; i++) {
		for (int j = t; j <= d; j++) {
			if (isdigit(matrix[i][j])){
				ret = matrix[i][j] - '0';
				if (++cnt > 1) return -cnt;
			}
		}
	}
	return ret;
}

void FillAns(int l, int t, int r, int d,char value) {
	for (int i = l; i <= r; i++) {
		for (int j = t; j <= d; j++) {
			ans[i][j] = value;
		}
	}
}

bool dfs(int index, char id) {
	//index一维的下标转化为二维下标  找空闲的
	while (ans[index / n][index % n] != '.') ++index;

	if (index == n * n) return true;

	//确定左上角坐标 x y
	int lx = index / n, ly = index % n,rBound = n,isover = 0;

	//找右下角坐标
	for (int rx = lx; rx < n; rx++) {
		for (int ry = ly; ry < rBound; ry++) {
			//向右扩展失败 当前已被填充 收缩右部
			if (ans[rx][ry] != '.') { rBound = ry;break; }
			//计算容量的人数
			int tot = (ry - ly + 1) * (rx - lx + 1);
			//太大了 人家不想要
			if (tot > 9) { rBound = ry; break; }
			//查看当前矩形局域是否只有一个小组且容量够
			int cnt = GetGroupCnt(lx, ly, rx, ry);
			
			//找到多个小组收缩 或者容量大了 
			if (cnt < 0 || (cnt > 0 &&  cnt < tot )) {
				//收缩右部
				rBound = ry;
				break;
			}

			//容量不够 继续扩展 或者一个都没有
			if(cnt > tot || cnt == 0) continue;

			//尝试填充
			FillAns(lx, ly, rx, ry, id);
			//能够找到解决方案就不回溯
			if (dfs(index, id + 1)) return true;
			//不行啊
			FillAns(lx, ly, rx, ry, '.');
		}
	}
	return false;
}

int main()
{
	while (cin >> n >> k && n && k) {
		memset(ans, '.', sizeof(ans));

		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				cin >> matrix[i][j];
			}
		}
		dfs(0, 'A');
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				cout << ans[i][j];
			}
			cout << endl;
		}
	}

	return 0;
}