HDu 1026 Ignatius and the Princess I(bfs検索+出力パス)

タイトル:hdu-1026 Ignatius and the Princess I Ignatius and the Princess I Time Limit:2000/1000 MS(Java/others)Memory Limit:65536/32768 K(Java/others)Total Submission(s):14677 Accepted Submission(s):4653 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. XXXXX1 5 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) FINISH It 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) FINISH God please help our poor hero. FINISH
文字型の迷路を入力して、左上から右下に歩いて、所要の最短时间とコースを求めて、一歩ごとに上、下、左、右しか歩けなくて、斜めに歩いてはいけなくて、図は'.'は歩いてもいいことを表して、しかも一歩ごとに时间が1秒かかることを表して、図が'数字'の时に数字の秒の时间が滞在することを表して、だからここを歩いて时间を费やします:map[][]-「0」+1;図が「X」であれば、歩いてはいけないことを示します.テーマ分析:この問題は迷宮検索で最短の問題を求め、出力は経路を含めて、典型的な検索問題でしょう.この問題はbfs検索を使用しています.まず、各ビットまで歩いた座標と費やした総時間を格納する構造体を構築する必要があります.bfsを使用するにはもちろん優先キューを使用します.このキューでは、時間のかかるものを先頭に置く.また、最短ルートを格納する構造体を構築し、最短パスを見つけたときに順次パスを出力すればよい.ACコード:

#include
#include
#include
#define MAX 110
using namespace std;
char map[MAX][MAX];
int mark[MAX][MAX];
int dx[4]={1,-1,0,0};               //  x         
int dy[4]={0,0,1,-1};               //  y         
struct node{
    int x,y;
    int time;
    friend bool operator < (node a,node b)
    {
        return a.time > b.time;         //    ，       
    }
};
struct Edge{
    int px;
    int py;                             //         
};
node edge[MAX];
Edge pre[MAX][MAX];
int n,m;

void bfs(int sx,int sy)
{
    mark[sx][sy]=1;         //        
    node now,next;          //       ，           
    pre[sx][sy].px=-1;      //     
    now.x=sx;
    now.y=sy;
    now.time=0;
    if(map[sx][sy]!='.')            
        now.time=map[sx][sy]-'0';
    priority_queueq;  //       
    q.push(now);
    while(!q.empty())
    {
        now=q.top();
        q.pop();
        if(now.x==0&&now.y==0)      //      ，       
        {
            printf("It takes %d seconds to reach the target position, let me show you the way.
",now.time);
            int key=1,total=now.time;
            sx=now.x,sy=now.y;
            while(pre[sx][sy].px!=-1)   //       ，       
            {
                int tx=pre[sx][sy].px;
                int ty=pre[sx][sy].py;
                printf("%ds:(%d,%d)->(%d,%d)
",key++,sx,sy,tx,ty);
                if(map[tx][ty]!='.')
                    for(int i=0;i<map[tx][ty]-'0';i++)
                        printf("%ds:FIGHT AT (%d,%d)
",key++,tx,ty);
                sx=tx;
                sy=ty;
            }
            return ;
        }
        for(int i=0;i<4;i++)    //       ，                   
        {
            next.x=now.x+dx[i];
            next.y=now.y+dy[i];
            if(next.x<0||next.x>=n)
                continue;
            if(next.y<0||next.y>=m)
                continue;
            if(mark[next.x][next.y])
                continue;
            mark[next.x][next.y]=1;
            if(map[next.x][next.y]=='X')
                continue;
            if(map[next.x][next.y]=='.')
                next.time=now.time+1;
            else
                next.time=now.time+map[next.x][next.y]-'0'+1;
            pre[next.x][next.y].px=now.x;   //        
            pre[next.x][next.y].py=now.y;
            q.push(next); 
        }
    }
    printf("God please help our poor hero.
");
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {   
        getchar();
        memset(mark,0,sizeof(mark));        //        
        for(int i=0;ifor(int j=0;j<m;j++)
                scanf("%c",&map[i][j]);     //    
            getchar();
        }
        bfs(n-1,m-1);                       //      ，                 
        printf("FINISH
");
    }
    return 0;
}