hdu 1026 (带路径BFS)

/*
题意：给个二维数组，'.'可以走，'X'不可走，'1-9'代表在此消耗的时间
输出记录从(0,0)到(n-1,m-1)的耗时最小值

YY:  ：从(n-1,m-1)到(0,0)逆向BFS，记录上一点，直接输出最优序列
*/


#include <iostream>
#include <queue>
using namespace std;
const int MAX = 99999999;

int n,m;
struct zu{
	char c;
	int num,x,y,qianx,qiany;
}map[110][110];
int dir[4][2]={{-1,0},{0,-1},{1,0},{0,1}};

void Print()
{
	int x, y, a, b, num, i;
	num = 1;
	if(map[0][0].num != MAX){
		printf("It takes %d seconds to reach the target position, let me show you the way.\n",map[0][0].num);
		
		x =0; y =0;
		while(x != n - 1 || y != m - 1){
			a = map[x][y].qianx;
			b = map[x][y].qiany;
			if(map[x][y].c >= '1' && map[x][y].c <= '9')
			{
				for(i = 0;i < map[x][y].c - '0'; ++i)
					printf("%ds:FIGHT AT (%d,%d)\n", num++, x, y);
			}
			printf("%ds:(%d,%d)->(%d,%d)\n", num++, x, y, a, b);
			x = a; y = b;
		}
		if(map[x][y].c >= '1' && map[x][y].c <= '9'){
			for(i = 0;i < map[x][y].c - '0'; ++i)
				printf("%ds:FIGHT AT (%d,%d)\n", num++, x, y);
		}
		printf("FINISH\n");
	}
	else{
		printf("God please help our poor hero.\nFINISH\n");
	}
}

void bfs()
{
	int i,bu;
	char temp_c;
	struct zu a;
	queue<struct zu> Q;
	//
    map[n-1][m-1].num = 0;
	
	if(map[n-1][m-1].c >= '1' && map[n-1][m-1].c <= '9')
		map[n-1][m-1].num = map[n-1][m-1].c - '0';
	else if(map[n-1][m-1].c == '.') 
		bu = 1;
	
	if(map[n-1][m-1].c != 'X')
		Q.push(map[n-1][m-1]);
	//
	while(!Q.empty())
	{
		a = Q.front();
		for(i = 0; i < 4; ++i)
		{
			int addx,addy;// 为后面代码简洁
			addx = dir[i][0];
			addy = dir[i][1];

			if(map[a.x + addx][a.y + addy].c == 'X' 
			   || a.y + addy >= m || a.x + addx < 0 || a.y + addy < 0 || a.x + addx >= n)
				continue;
			
			temp_c = map[a.x + addx][a.y + addy].c;
			if(temp_c == '.')		                	
				bu = a.num;
			else if(temp_c >= '1' && temp_c <= '9')		
				bu = a.num + temp_c - '0';  
			
			if(bu +1 < map[a.x + addx][a.y + addy].num) // 当出现比现在小的状态时更新
			{
				map[a.x + addx][a.y + addy].num = bu + 1;
				map[a.x + addx][a.y + addy].qianx = a.x;
				map[a.x + addx][a.y + addy].qiany = a.y;
				Q.push(map[a.x + addx][a.y + addy]);					
			}
		}
		Q.pop();
	}
	Print();
}
int main()
{
	int i,j;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		getchar();
		for(i = 0;i < n; ++i){
			for(j = 0;j < m; ++j){
				scanf("%c", &map[i][j].c);
				map[i][j].num = MAX;
				map[i][j].x = i;
				map[i][j].y = j;
				map[i][j].qianx = -1;
				map[i][j].qiany = -1;
			}
			getchar();
		}
		bfs();
	}
	return 0;
}