Poj 3083 Children of the Candy Corn

题目连接：http://poj.org/problem?id=1077

题目描述：给出一个迷宫，#表示墙，S表示起点，E表示终点。求从S到E的优先靠左边的墙走、优先靠右边的墙走和S到E的最短路径。

解法：求优先靠左靠右的路径，用dfs，求最短路径用bfs即可。

dfs中可以用nx = x+move[(i+4)%4][0];ny = y+move[(i+4)%4][1];处理很巧妙

View Code

#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<queue>
#define see(x) cout<<#x<<":"<<x<<endl;
using namespace std;
const int inf = 9999999;
const int N = 50;
char map

;
typedef struct{
int x, y, dis;
}Node;

const int move[4][4] ={{0,-1},{-1,0},{0,1},{1,0}};
bool vis

= {0};
int r, c;
queue<Node> q;
int bfs(){
int i;
Node temp, temp1;
int minn = inf;
while(!q.empty()){
temp = q.front();
q.pop();
if(map[temp.x][temp.y]=='E'){
if(temp.dis<minn){
minn = temp.dis;
}
}

else{
for(i=0;i<4;i++){
temp1.x = temp.x+move[i][0];
temp1.y = temp.y+move[i][1];
temp1.dis = temp.dis+1;
if(temp1.x>=0&&temp1.x<r&&temp1.y>=0&&temp1.y<c&&map[temp1.x][temp1.y]!='#'&&!vis[temp1.x][temp1.y]){
vis[temp1.x][temp1.y] = 1;
q.push(temp1);
}
}
}
}
return minn;
}
int dfs(int x, int y, int dir, char mark){
int i, ndir, nx, ny;
if(map[x][y]=='E'){
return 1;
}
if(mark=='l'){
for(i=dir-1;i<dir+3;i++){
nx = x+move[(i+4)%4][0];
ny = y+move[(i+4)%4][1];
ndir = (i+4)%4;
if(nx>=0&&nx<r&&ny>=0&&ny<c&&map[nx][ny]!='#'){
//    cout<<"nx: "<<nx<<"   ny: "<<ny<<"    "; see(ndir)
return dfs(nx,ny,ndir,mark)+1;
}
}
}
if(mark=='r'){
for(i=dir+1;i>dir-3;i--){
nx = x+move[(i+4)%4][0];
ny = y+move[(i+4)%4][1];
ndir = (i+4)%4;
if(nx>=0&&nx<r&&ny>=0&&ny<c&&map[nx][ny]!='#'){
return dfs(nx,ny,ndir,mark)+1;
}
}
}
return 0;
}
int main(){
int t, lp, rp, sp;
int i, j, k, dir;
Node s;
scanf("%d",&t);
while(t--){
scanf("%d%d",&c,&r);
for(i=0;i<r;i++)
scanf("%s",map[i]);
for(i=0;i<r;i++){
for(j=0;j<c;j++){
if(map[i][j]=='S'){
s.x=i; s.y=j; s.dis=1;
break;
}
}
}
if(s.x==0){
dir=3;
}
else if(s.x==1){
dir=1;
}
else if(s.y==0){
dir = 2;
}
else{
dir = 0;
}
lp = dfs(s.x,s.y,dir,'l');

vis[s.x][s.y] = 1;
rp = dfs(s.x,s.y,dir,'r');

memset(vis,0,sizeof(vis));
vis[s.x][s.y] = 1;  s.dis = 1;
q.push(s);
sp = bfs();
printf("%d %d %d\n",lp,rp,sp);
}
return 0;
}