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2013
12-11

HDU 1498 50 years, 50 colors-DFS-[解题报告] C++

50 years, 50 colors

问题描述 :

On Octorber 21st, HDU 50-year-celebration, 50-color balloons floating around the campus, it’s so nice, isn’t it? To celebrate this meaningful day, the ACM team of HDU hold some fuuny games. Especially, there will be a game named "crashing color balloons".

There will be a n*n matrix board on the ground, and each grid will have a color balloon in it.And the color of the ballon will be in the range of [1, 50].After the referee shouts "go!",you can begin to crash the balloons.Every time you can only choose one kind of balloon to crash, we define that the two balloons with the same color belong to the same kind.What’s more, each time you can only choose a single row or column of balloon, and crash the balloons that with the color you had chosen. Of course, a lot of students are waiting to play this game, so we just give every student k times to crash the balloons.

Here comes the problem: which kind of balloon is impossible to be all crashed by a student in k times.

输入:

There will be multiple input cases.Each test case begins with two integers n, k. n is the number of rows and columns of the balloons (1 <= n <= 100), and k is the times that ginving to each student(0 < k <= n).Follow a matrix A of n*n, where Aij denote the color of the ballon in the i row, j column.Input ends with n = k = 0.

输出:

For each test case, print in ascending order all the colors of which are impossible to be crashed by a student in k times. If there is no choice, print "-1".

样例输入:

1 1
1
2 1
1 1
1 2
2 1
1 2
2 2
5 4
1 2 3 4 5
2 3 4 5 1
3 4 5 1 2
4 5 1 2 3
5 1 2 3 4
3 3
50 50 50
50 50 50
50 50 50
0 0

样例输出:

-1
1
2
1 2 3 4 5
-1

/*
题意:n*n的矩形中放入颜色值为[1,50]的气球,要求每一个人扎k次,每扎一次,可以将同行或者同列相同颜色的气球全部扎破。求是否存在不可能全部扎破的气球,按照升序规律输出气球的颜色。

题解:多次进行最小点覆盖运算即可
矩形行列分别为集合A和集合B,如果判断k气球,则如果map[i][j] = k,则表示存在一条边,这样便可以转换成最小点覆盖问题,只需要找出最小的点,清除掉两集合之间所有的边即可。
*/

#include <iostream>
#define re(i, n) for(int i = 0; i < n; ++ i)
using namespace std;

const int nMax = 105;
int map[nMax][nMax];
int link[nMax];
int useif[nMax];
int ans[nMax];
int len;
int n, k;

int dfs(int t, int col)
{
	re(i, n)
	{
		if(!useif[i] && map[t][i] == col)
		{
			useif[i] = 1;
			if(link[i] == -1 || dfs(link[i], col))
			{
				link[i] = t;
				return 1;
			}
		}
	}
	return 0;
}

int maxMatch(int col)
{
	memset(link, -1, sizeof(link));
	int num = 0;
	re(i, n)
	{
		memset(useif, 0, sizeof(useif));
		if(dfs(i, col)) num ++;
	}
	return num;
}

int main()
{
	//freopen("f://data.in", "r", stdin);
	while(scanf("%d %d", &n, &k) != EOF)
	{
		memset(map, 0, sizeof(map));
		len = 0;
		if(!n && !k) break;
		re(i, n) re(j, n) scanf("%d", &map[i][j]);
		for(int i = 1; i <= 50; ++ i)
		{
			if(maxMatch(i) > k)
				ans[len ++] = i;
		}
		if(!len) 
			printf("-1\n");
		else
		{
			re(i, len - 1) printf("%d ", ans[i]);
			printf("%d\n",ans[len - 1]);
		}
	}
	return 0;
}

解题报告转自:http://blog.csdn.net/lhshaoren/article/details/7761222


  1. 5.1处,反了;“上一个操作符的优先级比操作符ch的优先级大,或栈是空的就入栈。”如代码所述,应为“上一个操作符的优先级比操作符ch的优先级小,或栈是空的就入栈。”

  2. 我还有个问题想请教一下,就是感觉对于新手来说,递归理解起来有些困难,不知有没有什么好的方法或者什么好的建议?