Hdu 1026 – Ignatius and the Princess

Problem DescriptionThe 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.InputThe 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.OutputFor 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

算是一道比較難的搜尋題了吧!參考海潮哥的代碼....

#include<iostream>using namespace std;#define maxx 10010char map[110][110];int vis[110][110],f[110][110];struct node{    int x,y,t;    node(){}    node(int a,int b,int c):x(a),y(b),t(c){} }path[maxx],que[maxx*9];int n,m;int fx[]={1,-1,0,0};int fy[]={0,0,1,-1};int bfs(){        int i,j,t,step;        memset(vis,-1,sizeof(vis));        for(i=0;i<n;i++){            scanf("%s",map[i]);            for(j=0;j<m;j++)            if(map[i][j]=='.')  map[i][j]='0';            }        int s,e;        que[s=e=0]=node(0,0,map[0][0]-'0');        vis[0][0]=0;        while(s<=e)           {            node no=que[s++];            if(no.t){// 原地打怪，停留在原地打                 que[++e]=node(no.x,no.y,no.t-1);                vis[no.x][no.y]++;                continue;            }            if(n-1==no.x && m-1==no.y) break;//到達終點跳出             for(int i=0;i<4;i++){                int nx=no.x+fx[i],ny=no.y+fy[i];                if(nx<0||ny<0||nx>=n||ny>=m) continue;                if(map[nx][ny]=='X'||vis[nx][ny]!=-1) continue;                vis[nx][ny]=vis[no.x][no.y]+1;                f[nx][ny]=i;//標記方向                 que[++e]=node(nx,ny,map[nx][ny]-'0');            }        }        return vis[n-1][m-1];}int main(){    int x,y,nx,ny,i,j,tol;    while(~scanf("%d%d",&n,&m))    {    tol=bfs();    if(tol==-1)        puts("God please help our poor hero.");    else{        int step=0;        x=n-1,y=m-1;        while(x||y){// 通過f[][]記錄的方向，我們可以知道是從哪個座標走到這一個座標的，通過這個方式將路線匯出到path                path[step++]=node(x,y,map[x][y]-'0');                nx=x+fx[f[x][y]^1],ny=y+fy[f[x][y]^1];                x=nx,y=ny;        }        path[step]=node(0,0,map[0][0]-'0');        printf("It takes %d seconds to reach the target position, let me show you the way.\n",tol);        int t=0;        while(step--){            printf("%ds:(%d,%d)->(%d,%d)\n",++t,path[step+1].x,path[step+1].y,path[step].x,path[step].y);            while(path[step].t--) printf("%ds:FIGHT AT (%d,%d)\n",++t,path[step].x,path[step].y);                }                    }    puts("FINISH");        }    return 0;}