题意：机器人在网格线上行走，从p1点开始，沿最短路径到p2，再沿最短路径到p3，依此类推。在此过程中留下了行走的运动轨迹，由“RLDU”表示。问若只给出运动轨迹，求最少的pi点的个数。
分析：pi到pi+1是沿最短路径走的，因此在此路径中不可能同时出现“UD”两个方向（“LR”同理）。因此只要同时出现，那一定证明此刻已往下一个目标点走。
#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b){
if(fabs(a - b) < eps) return 0;
return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 200000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
char a[MAXN];
map<char, int> mp;
int vis[5];
void init(){
mp['L'] = 0;
mp['R'] = 1;
mp['U'] = 2;
mp['D'] = 3;
}
bool judge(int x){
if(x == 0 && vis[1]) return true;
if(x == 1 && vis[0]) return true;
if(x == 2 && vis[3]) return true;
if(x == 3 && vis[2]) return true;
return false;
}
int main(){
init();
int n;
while(scanf("%d", &n) == 1){
scanf("%s", a);
int cnt = 1;
memset(vis, 0, sizeof vis);
for(int i = 0; i < n; ++i){
int tmp = mp[a[i]];
if(judge(tmp)){
memset(vis, 0, sizeof vis);
++cnt;
}
vis[tmp] = 1;
}
printf("%d\n", cnt);
}
return 0;
}