hdu1026 Ignatius and the Princess I

来源：互联网发布：淘宝女装模特拍摄技巧编辑：程序博客网时间：2024/05/16 02:00

http://acm.hdu.edu.cn/showproblem.php?pid=1026

题目大意：

就是搜索算法加上保存路径。

用一个数组来保存每一次移动时的方向，最后再从终点向起点递归输出。

#include<iostream>#include<stdio.h>#include<string.h>#include<queue>using namespace std;int n,m;int a[4][2]={{1,0},{-1,0},{0,-1},{0,1}};struct node{int x,y;int step;friend bool operator<(node x,node y){return x.step>y.step;}};int map[105][105];int blood[105][105];int flag[105][105];int bfs(){int i,j,xx,yy;node cur,next;priority_queue<node>q;cur.x=0;cur.y=0;cur.step=0;q.push(cur);while(!q.empty()){cur=q.top();q.pop();if(cur.x==n-1&&cur.y==m-1)return cur.step;for(i=0;i<4;i++){next=cur;next.x=cur.x+a[i][0];next.y=cur.y+a[i][1];if(next.x>=0&&next.x<n&&next.y>=0&&next.y<m&&map[next.x][next.y]>0){next.step=cur.step+map[next.x][next.y];map[next.x][next.y]=-1;flag[next.x][next.y]=i;//printf("i=%d %d %d %d %d\n",i,cur.x,cur.y,next.x,next.y);q.push(next);}}}return -1;}int tmp;void p(int x,int y){int xx,yy;if(flag[x][y]==-1)return;xx=x-a[flag[x][y]][0];yy=y-a[flag[x][y]][1];//printf("%d %d\n",xx,yy);p(xx,yy);printf("%ds:(%d,%d)->(%d,%d)\n",tmp++,xx,yy,x,y);while(blood[x][y]--)printf("%ds:FIGHT AT (%d,%d)\n",tmp++,x,y);}int main(){char st[105];while(scanf("%d%d",&n,&m)!=EOF){int i,j;int ans;memset(flag,-1,sizeof(flag));memset(map,-1,sizeof(map));memset(blood,0,sizeof(blood));for(i=0;i<n;i++){scanf("%s",st);for(j=0;j<m;j++){if(st[j]=='X')map[i][j]=-1;else if(st[j]=='.')map[i][j]=1;else{map[i][j]=st[j]-'0'+1;blood[i][j]=st[j]-'0';}//printf("%d",map[i][j]);}//printf("\n");}map[0][0]=-1;ans=bfs();if(ans==-1){printf("God please help our poor hero.\n");}else{tmp=1;//printf("%d\n",flag[n-1][m-1]);//printf("0 0 %d\n",flag[0][0]);printf("It takes %d seconds to reach the target position, let me show you the way.\n",ans);p(n-1,m-1);}printf("FINISH\n");}}

0 0