Driving Straight
Problem Description
The city where Peter lives has the form of the rectangle with M streets running from east to west and N streets running from north to south. Recently, because of preparations for the celebration of the city’s 3141-st birthday, some street sectors has been closed for driving.
Peter lives in the house next to point (1, 1) and works next to the point (N, M ). He always drives from home to work by his car using the shortest possible path. Of course, he can’t drive through closed sectors. Since there can be many shortest pathes between his house and his work, he may choose any.
But Peter doesn’t like to turn (he is an inexperienced driver), so he wants to choose the path using the following algorithm: starting from the point (1, 1) he drives either northwards, or eastwards (wherever there is the shortest path available, if there are both, he may choose any). Whenever he comes to the junction he must decide where to go. If there is only one direction he can drive to stay on the shortest path, he must choose that direction. In the other case he would like to choose the direction giving priority to driving forward, that is, if he can drive forward and still stay on some shortest path, he would go forward. If he can’t drive forward to stay on the shortest path, he would choose any available direction.
Help Peter to create the path from his house to his work using the rules described.
Input
Next 2M − 1 lines contain 2N − 1 characters each, representing the city map. House blocks are marked with spaces, junctions with ‘+’, open sectors with ‘-’ and ’|’, closed sectors with spaces. Peter’s house is at the lower-left corner of the map, his work is at the upper-right corner.
Output
Sample Input
4 4 +-+ +-+ | | | + +-+-+ | | +-+-+-+ | | +-+-+-+
Sample Output
N RFLRL
Source
廣搜題目，從終點逆推，每個點每個方向是否是最短路都可以標記出來，有不少細節需要自己處理好，比如矩陣化圖，標記已經搜過的點等。總的來說就是一道水題。
代碼：
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<ctime>
#include<string>
#include<cstring>
#include<algorithm>
#include<fstream>
#include<queue>
#include<stack>
#include<vector>
#include<cmath>
#include<iomanip>
#define rep(i,n) for(i=1;i<=n;i++)
#define MM(a,t) memset(a,t,sizeof(a))
#define INF 1e9
typedef long long ll;
#define mod 1000000007
using namespace std;
int edge[700000],nxt[700000],fir[200000],lst[200000];
int n,m,n2,ST,TR,vis[200000];
bool f[200000][4];
queue<int> qu;
void addedge(int s,int e){
edge[++n2]=e;
if(fir[s]==-1){fir[s]=n2; lst[s]=n2;}
else {nxt[lst[s]]=n2; lst[s]=n2;}
}
int getnode(int x,int y){
x=(x+1)/2;
y=(y+1)/2;
return (x-1)*m+y;
}
int getdir(int s,int e){
if(s==e+1) return 0;
if(s==e+m) return 1;
if(s==e-1) return 2;
if(s==e-m) return 3;
}
int getnext(int x,int dir){
switch(dir){
case 0:return x-1;
case 1:return x-m;
case 2:return x+1;
case 3:return x+m;
}
}
int main()
{
int i,j,dir,len;
char st[1000];
while(scanf("%d%d",&n,&m)!=EOF){
n2=0; MM(fir,-1); MM(nxt,-1); MM(lst,-1); MM(vis,0);
ST=getnode(2*n-1,1); TR=getnode(1,2*m-1); MM(f,0);
getchar();
for(i=1;i<=2*n-1;i++){
cin.getline(st,2*m);
len=strlen(st);
if(len!=2*m-1){
for(j=len;j<2*m-1;j++) st[j]=' ';
st[2*m-1]='\0';
}
if(i%2){
int s,e;
for(j=1;j<=2*(m-1)-1;j+=2)
if(st[j]!=' '){
s=getnode(i,j); e=getnode(i,j+2);
addedge(s,e);
addedge(e,s);
}
}
else{
int s,e;
for(j=0;j<=2*(m-1);j+=2)
if(st[j]!=' '){
s=getnode(i-1,j+1); e=getnode(i+1,j+1);
addedge(s,e);
addedge(e,s);
}
}
}//作圖
while(!qu.empty()) qu.pop();
qu.push(TR);
vis[TR]=1;
while(!qu.empty()){
int s,e=qu.front(); qu.pop();
if(e==ST) break;
for(i=fir[e];i!=-1;i=nxt[i])
{
s=edge[i];
if(vis[s]==2) continue;
f[s][getdir(s,e)]=1;
if(!vis[s]){
vis[s]=1;
qu.push(s);
}
}
vis[e]=2;
}
if(f[ST][1]){
printf("N\n");
ST-=m; dir=1;
}
else{
printf("E\n");
ST+=1; dir=2;
}
while(ST!=TR){
if(f[ST][dir]){
printf("F");
ST=getnext(ST,dir);
continue;
}
int l=(dir+3)%4,r=(dir+5)%4;
if(f[ST][l]){
printf("L");
dir=l;
ST=getnext(ST,l);
}
else{
printf("R");
dir=r;
ST=getnext(ST,r);
}
}
printf("\n");
}
return 0;
}其實碼完才發現不用化圖，本來化圖是爲了可能可以套什麼模板，沒想到直接可以做了。