hdu 1026 Ignatius and the Princess I bfs+优先级队列

Ignatius and the Princess I

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10388    Accepted Submission(s): 3146
Special Judge

Problem Description

The Princess has been abducted by the BEelzebub feng5166, our hero Ignatius has to rescue our pretty Princess. Now he gets into feng5166's castle. The castle is a large labyrinth. To make the problem simply, we assume the labyrinth is a N*M two-dimensional array which left-top corner is (0,0) and right-bottom corner is (N-1,M-1). Ignatius enters at (0,0), and the door to feng5166's room is at (N-1,M-1), that is our target. There are some monsters in the castle, if Ignatius meet them, he has to kill them. Here is some rules:

1.Ignatius can only move in four directions(up, down, left, right), one step per second. A step is defined as follow: if current position is (x,y), after a step, Ignatius can only stand on (x-1,y), (x+1,y), (x,y-1) or (x,y+1).
2.The array is marked with some characters and numbers. We define them like this:
. : The place where Ignatius can walk on.
X : The place is a trap, Ignatius should not walk on it.
n : Here is a monster with n HP(1<=n<=9), if Ignatius walk on it, it takes him n seconds to kill the monster.

Your task is to give out the path which costs minimum seconds for Ignatius to reach target position. You may assume that the start position and the target position will never be a trap, and there will never be a monster at the start position.

 

 

Input

The input contains several test cases. Each test case starts with a line contains two numbers N and M(2<=N<=100,2<=M<=100) which indicate the size of the labyrinth. Then a N*M two-dimensional array follows, which describe the whole labyrinth. The input is terminated by the end of file. More details in the Sample Input.

 

 

Output

For each test case, you should output "God please help our poor hero." if Ignatius can't reach the target position, or you should output "It takes n seconds to reach the target position, let me show you the way."(n is the minimum seconds), and tell our hero the whole path. Output a line contains "FINISH" after each test case. If there are more than one path, any one is OK in this problem. More details in the Sample Output.

 

 

Sample Input

5 6.XX.1...X.2.2...X....XX.XXXXX.5 6.XX.1...X.2.2...X....XX.XXXXX15 6.XX.....XX1.2...X....XX.XXXXX.

 

 

Sample Output

It takes 13 seconds to reach the target position, let me show you the way.1s:(0,0)->(1,0)2s:(1,0)->(1,1)3s:(1,1)->(2,1)4s:(2,1)->(2,2)5s:(2,2)->(2,3)6s:(2,3)->(1,3)7s:(1,3)->(1,4)8s:FIGHT AT (1,4)9s:FIGHT AT (1,4)10s:(1,4)->(1,5)11s:(1,5)->(2,5)12s:(2,5)->(3,5)13s:(3,5)->(4,5)FINISHIt takes 14 seconds to reach the target position, let me show you the way.1s:(0,0)->(1,0)2s:(1,0)->(1,1)3s:(1,1)->(2,1)4s:(2,1)->(2,2)5s:(2,2)->(2,3)6s:(2,3)->(1,3)7s:(1,3)->(1,4)8s:FIGHT AT (1,4)9s:FIGHT AT (1,4)10s:(1,4)->(1,5)11s:(1,5)->(2,5)12s:(2,5)->(3,5)13s:(3,5)->(4,5)14s:FIGHT AT (4,5)FINISHGod please help our poor hero.FINISH

 

 

第一眼看到题目 天真的以为 暴力DFS就可以了。。。后来看了代码 才发现用的是bfs。。  自己之前没接触过优先级队列。。顿时觉得自己是村儿里人了。。

优先级队列。。  是个好东西。。。 那这题就没什么难度了。。。我调试了好久才发现出现问题了地方。。

（STL是个好东西 多用用不错。。）

 

#include<iostream>#include<cstring>#include<cstdio>#include<string>#include<algorithm>#include<queue>using namespace std;int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};//下上右左char g[110][110];bool vis[110][110];int n,m;struct node{    int x,y;    int t;    int id,pre;    bool operator <(const node &a)const    {        return t>a.t;    }}que[10010];int en;bool jud(int x,int y){    if(x<0||x>=n) return false;    if(y<0||y>=m) return false;    if(vis[x][y]||g[x][y]=='X')        return false;    return true;}int bfs(){    int x,y;    priority_queue<node> q;    node a;    en=0;    que[en].x=0,que[en].y=-1,que[en].t=-1;    que[en].id=0,que[en].pre=-1;    q.push(que[en++]);    memset(vis,0,sizeof(vis));    while(!q.empty())    {        a=q.top();        q.pop();        for(int i=0;i<4;i++)        {            x=a.x+dir[i][0];            y=a.y+dir[i][1];            if(jud(x,y))            {                que[en].x=x;                que[en].y=y;                que[en].id=en;                que[en].t=a.t+(g[x][y]=='.'?1:g[x][y]-'0'+1);                que[en].pre=a.id;                if(x==n-1 && y==m-1)                    return en;                q.push(que[en++]);                vis[x][y]=1;            }        }    }    return -1;}void out(int id){    int pre=que[id].pre;    int x=que[id].x;    int y=que[id].y;    int px=que[pre].x;    int py=que[pre].y;    if(que[id].pre==1)    {        printf("%ds:(%d,%d)->(%d,%d)\n",que[pre].t+1,px,py,x,y);        if(g[x][y]!='.')        {            int t=g[x][y]-'0';            for(int i=1;i<=t;i++)                printf("%ds:FIGHT AT (%d,%d)\n",que[pre].t+i+1,x,y);        }    }    else    {        out(pre);        printf("%ds:(%d,%d)->(%d,%d)\n",que[pre].t+1,px,py,x,y);        if(g[x][y]!='.')        {            int t=g[x][y]-'0';            for(int i=1;i<=t;i++)                printf("%ds:FIGHT AT (%d,%d)\n",que[pre].t+i+1,x,y);        }    }}int main(){    while(~scanf("%d%d",&n,&m))    {        gets(g[0]);        for(int i=0;i<n;i++)            gets(g[i]);        int id=bfs();        if(id!=-1)        {            printf("It takes %d seconds to reach the target position, let me show you the way.\n",que[id].t);            out(id);            printf("FINISH\n");        }        else            printf("God please help our poor hero.\nFINISH\n");    }    return 0;}