Robot Cleaner I
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=4092
C++版本一
题解：
1、有个东西叫%1d；
2、哈希坐标；
3、剪枝；
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
//#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=2000+100;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m;
ll k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int a,b,mp[N][N];
int vis[N][300];
char str[N],str1[N];
struct node{
int x,y,step;
}tmp,start,front1;
int command(int x,int y){
return 81*mp[x][y]+27*mp[x-1][y]+9*mp[x+1][y]+3*mp[x][y-1]+mp[x][y+1];
}
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
//ios::sync_with_stdio(false);
//cin.tie(0);
//cout.tie(0);
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m);
scanf("%d%d%lld",&a,&b,&k);
scanf("%s",str);
sum=0;
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
scanf("%1d",&mp[i][j]);
if(mp[i][j]==2)
sum++;
}
}
start.x=a;
start.y=b;
start.step=0;
memset(vis,-1,sizeof(vis));
ans=0;
front1=start;
while(1){
int opid=command(front1.x,front1.y);
char op=str[opid];
if(vis[(front1.x-1)*m+front1.y][opid]==ans)
break;
if(sum<=ans)
break;
vis[(front1.x-1)*m+front1.y][opid]=ans;
tmp=front1;
tmp.step++;
if(tmp.step>k)
break;
//cout<<front1.x<<" "<<front1.y<<" "<<op<<" "<<mp[front1.x][front1.y]<<" "<<vis[front1.x][front1.y]<<endl;
if(op=='U'){
tmp.x--;
}else if(op=='D'){
tmp.x++;
}else if(op=='L'){
tmp.y--;
}else if(op=='R'){
tmp.y++;
}else if(op=='P'){
if(mp[front1.x][front1.y]==2){
mp[front1.x][front1.y]=0;
ans++;
}
}else if(op=='I'){
break;
}
if(mp[tmp.x][tmp.y]==1){
break;
}
front1=tmp;
}
printf("%d\n",ans);
}
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Hello world!" << endl;
return 0;
}
C++版本二
题解：注意标记，vis[i][j] i表示对坐标的哈希，j表示当前计算的x，值为之前得到多少汽油了，因为如果再次经过这个位置，两次经过的中间得到多汽油，并且改变了图中的2，所以要继续走，知道，走到一个位置，汽油数不增加，然后就是注意一些剪枝。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n,m;
int a,b;
ll k;
char mp[2010][2010];
int vis[2100][255];
char step[255];
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
scanf("%d%d%lld",&a,&b,&k);
scanf("%s",step+1);
int sum=0;
for(int i=0;i<=n*m;i++)
for(int j=0;j<=250;j++)
vis[i][j]=-1;
for(int i=1;i<=n;i++)
{
scanf("%s",mp[i]+1);
for(int j=1;j<=m;j++)
{
if(mp[i][j]=='2')
sum++;
}
}
int ans=0;
int x;
int xx,yy;
int st=0;
while(1)
{
x=81*(mp[a][b]-'0')+27*(mp[a-1][b]-'0')+9*(mp[a+1][b]-'0')+3*(mp[a][b-1]-'0')+(mp[a][b+1]-'0');
if(vis[(a-1)*m+b][x]==ans) break;
vis[(a-1)*m+b][x]=ans;
x++;
// cout<<a<<" "<<b<<" "<<x<<" "<<step[x]<<" "<<endl;
if(step[x]=='U') xx=a-1,yy=b;
if(step[x]=='D') xx=a+1,yy=b;
if(step[x]=='L') xx=a,yy=b-1;
if(step[x]=='R') xx=a,yy=b+1;
if(step[x]=='P')
{
if(mp[a][b]=='2')
{
ans++;
mp[a][b]='0';
}
if(ans>=sum) break;
xx=a,yy=b;
}
if(step[x]=='I')
{
xx=a,yy=b;
break;
}
if(mp[xx][yy]=='1') break;
st++;
if(st==k) break;
a=xx;
b=yy;
}
printf("%d\n",ans);
}
return 0;
}
