hdu 1026 Ignatius and the Princess I(广搜+优先队列)

/*

题意：输出点（0，0）到（n-1，m-1）的最短路径。有数字的地方表示停留的时间。

解题：简单的广搜，就是多了一个路径的保存。在搜素的过程中用标记数组mark表示是否经过，顺便保存方向

*/

#include<iostream>
#include<string>
#include<queue>
#include<cstring>
using namespace std;
#define inf = 0xffffff
const int N=110;
int n, m, mark[N][N], time, dir[4][2]={1,0, 0,1 ,-1,0, 0,-1}, step;
char str[N][N];

struct node
{
	int x, y, step;
	friend bool operator < (node n1, node n2)
	{
		return n2.step < n1.step;
	}
};

int judge(int x, int y)
{
	if(x>=0 && x<n && y>=0 && y<m && mark[x][y]==0 && str[x][y] != 'X')
		return 1;
	else
		return 0;
}

int bfs(int x, int y)
{
	priority_queue<node>q;
	node cur, next;
	int i;
	cur.x = n-1; cur.y = m-1;; cur.step=0;
	q.push(cur);
	while(!q.empty())
	{
		cur=q.top();
		q.pop();
		if(cur.x==0 && cur.y==0) return cur.step;
		for( i=0; i < 4; i++ )
		{
			next.x = x = cur.x+dir[i][0];
			next.y = y = cur.y+dir[i][1];
			int t=0;
			if(str[x][y]>'0' && str[x][y]<='9')
				t = str[x][y]-'0';
			next.step = cur.step+t+1;
			if(judge(x, y))
			{
				mark[x][y]=i+1;
				q.push(next);
			}
		}
	}
	return -1;
}

void print(int x, int y)
{
	int next_x, next_y;
	if(x==n-1 && y==m-1) return ;
	next_x = x - dir[mark[x][y]-1][0];
	next_y = y - dir[mark[x][y]-1][1];
	int t = str[next_x][next_y] -'0';
	printf("%ds:(%d,%d)->(%d,%d)\n", time++, x,y, next_x,next_y);
	if(t>0 && t<=9)
	{
		int i;
		for( i=0; i < t; i++)
			printf("%ds:FIGHT AT (%d,%d)\n", time++, next_x, next_y);
	}
	print(next_x, next_y);
}

int main()
{
	int i, step;
	while(scanf("%d%d", &n, &m) != EOF )
	{
		step = 0;
		for( i=0; i < n; i++ )
			scanf("%s", str[i]);
		memset(mark, 0, sizeof(mark));
		mark[n-1][m-1]=5;
		step = bfs(n-1, m-1);
		if(step==-1)
			printf("God please help our poor hero.\n");
		else
		{
			time=1;
			if(str[n-1][m-1]>'0' && str[n-1][m-1]<='9' && n!=1 && m!=1)
				step += str[n-1][m-1]-'0';
			if(str[0][0]>'0' && str[0][0]<='9' && n!=1 && m!=1)
				step -= str[0][0]-'0';
			printf("It takes %d seconds to reach the target position, let me show you the way.\n", step);
			print(0, 0);
		}
		printf("FINISH\n");
	}
	return 0;
}