Ignatius and the Princess I

给你一个图，图上存在一些monster，每一次打死一个怪，要消耗一定的时间值。求最短时间到达终点！很简答的一个宽搜题。由于打怪耗时，所以采用延迟搜索！题目的关键是要输出路径，采用dfs深度优先！http://acm.hdu.edu.cn/showproblem.php?pid=1026

#include <iostream>#include <cstdio>#include <string.h>#include <queue>#define N 101using namespace std;int map[N][N],f[N][N];bool v[N][N];int jump[4][2]={{1,0},{-1,0},{0,1},{0,-1}};int n,m,res,ks;struct node{    int x,y,s;}q,p,pre[N][N];void init(){    int i,j;    char c;res=-1;ks=1;memset(map,0,sizeof(map));memset(f,0,sizeof(f));    for(i=0;i<n;i++)    {getchar();        for(j=0;j<m;j++)        {            scanf("%c",&c);            if(c=='.') map[i][j]=-1;            else if(c=='X') map[i][j]=0;            else map[i][j]=c-48;f[i][j]=map[i][j];        }    }memset(v,0,sizeof(v));    memset(pre,0,sizeof(pre));}int is_b(node p){    if(p.x<0 || p.y<0 || p.x>=n || p.y>=m||map[p.x][p.y]==0 || v[p.x][p.y])    return 1;    return 0;}void bfs(){    int i;    queue< node > que;    q.x=q.y=q.s=0;v[0][0]=1;map[0][0]=0;if(f[0][0]>=1) q.s=f[0][0];//考虑起点有怪    que.push(q);    while(!que.empty())    {        q=que.front();que.pop();if(q.x==n-1 && q.y==m-1 && map[q.x][q.y]==-1){res=q.s;//时间okreturn;}if(map[q.x][q.y]>=1){++q.s;if(map[q.x][q.y]==1) map[q.x][q.y]=-1;else map[q.x][q.y]-=1;que.push(q);}else {for(i=0;i<4;i++){p.x=q.x+jump[i][0];p.y=q.y+jump[i][1];p.s=q.s+1;if(is_b(p)) continue;v[p.x][p.y]=1;pre[p.x][p.y].x=q.x;pre[p.x][p.y].y=q.y;que.push(p);}}    }}void dfs(int x1,int y1){if(x1==0 && y1==0) return; dfs(pre[x1][y1].x , pre[x1][y1].y);int w=pre[x1][y1].x,e=pre[x1][y1].y;if(f[w][e]>=1){for(int i=1;i<=f[w][e];i++){printf("%ds:FIGHT AT (%d,%d)\n",ks++,w,e);}}printf("%ds:(%d,%d)->(%d,%d)\n",ks++,w,e,x1,y1);}void output(){//测试函数for(int i=0;i<n;i++){for(int j=0;j<m;j++)printf("%d",f[i][j]);puts("");}}int main (){//freopen("k.txt","r",stdin);int i;    while(~scanf("%d%d",&n,&m))    {        init();//output();        bfs();if(res==-1){printf("God please help our poor hero.\n");}else {printf("It takes %d seconds to reach the target position, let me show you the way.\n",res);dfs(n-1,m-1);if(f[n-1][m-1]>=1){for(i=0;i<f[n-1][m-1];i++)printf("%ds:FIGHT AT (%d,%d)\n",ks++,n-1,m-1);}}printf("FINISH\n");    }    return 0;}