HDU 1035 Robot Motion Union Find 并查集题解

本题的类型我一看就想到使用并查集解了，因为要查找是否有环，这是并查集的典型用法。

但是由于本题数据实在是太水了，故此有人使用直接模拟都过了。这让本题降了个档次。

这里使用并查集解。而且可以根据需要简化并查集函数，代码还是很好打的。

#include <stdio.h>
#include <vector>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <string>
#include <limits.h>
#include <stack>
#include <queue>
#include <set>
#include <map>
using namespace std;

const int MAX_N = 11;
const int MAX_V = MAX_N * MAX_N;

struct Subset
{
	int step, p;
};

Subset sub[MAX_V];
char Maze[MAX_N][MAX_N];
int R, C, sx = 0, sy;

int findPar(int x)
{
	if (sub[x].p) return sub[x].p = findPar(sub[x].p);
	return x;
}

inline void unionTwo(int x, int p)
{
	sub[x].p = p;
}

inline bool isLegal(int r, int c)
{
	return 0<=r && 0 <= c && r<R && c<C;
}

bool exitMaze(int &out, int &loop)
{
	int step = 0, si = sx, sj = sy, sni, snj;
	memset(sub, 0, sizeof(Subset)*(R*C+1));
	while (true)
	{
		step++;
		int p = si * C + sj + 1;//注意下标0不使用，作为空标志
		sub[p].step = step;

		//下一个格
		if (Maze[si][sj] == 'S')
		{
			sni = si+1, snj = sj;
		}
		else if (Maze[si][sj] == 'W')
		{
			sni = si, snj = sj-1;
		}
		else if (Maze[si][sj] == 'E')
		{
			sni = si, snj = sj+1;
		}
		else
		{
			sni = si-1, snj = sj;
		}

		if (!isLegal(sni, snj))//out of one side of Maze
		{
			out = step;
			return true;
		}

		int son = sni * C + snj + 1;//注意0不使用

		int pf = findPar(p);
		int sonf = findPar(son);

		if (pf == sonf)
		{
			out = sub[son].step-1;
			loop = step - out;
			return false;
		}

		unionTwo(son, pf);

		si = sni, sj = snj;
	}
}

int main()
{
	while (~scanf("%d %d %d", &R, &C, &sy) && R)
	{
		getchar();
		for (int i = 0; i < R; i++)
		{
			gets(Maze[i]);
		}
		sy--;
		int out = 0, loop = 0;
		if (exitMaze(out, loop))
		{
			printf("%d step(s) to exit\n", out);
		}
		else
		{
			printf("%d step(s) before a loop of %d step(s)\n", out, loop);
		}
	}
	return 0;
}