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

2014-02-01 23:03 453 查看
/*

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

, time, dir[4][2]={1,0, 0,1 ,-1,0, 0,-1}, step;
char str

;

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;
}
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