Codeforces Round #389 (Div. 2) C. Santa Claus and Robot 贪心+最短路

C. Santa Claus and Robot

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Santa Claus has Robot which lives on the infinite grid and can move along its lines. He can also, having a sequence of m pointsp1, p2, ..., pm with integer coordinates, do the following: denote its initial location by p0. First, the robot will move from p0 to p1 along one of the shortest paths between them (please notice that since the robot moves only along the grid lines, there can be several shortest paths). Then, after it reaches p1, it'll move to p2, again, choosing one of the shortest ways, then to p3, and so on, until he has visited all points in the given order. Some of the points in the sequence may coincide, in that case Robot will visit that point several times according to the sequence order.

While Santa was away, someone gave a sequence of points to Robot. This sequence is now lost, but Robot saved the protocol of its unit movements. Please, find the minimum possible length of the sequence.

Input

The first line of input contains the only positive integer n (1 ≤ n ≤ 2·105) which equals the number of unit segments the robot traveled. The second line contains the movements protocol, which consists of n letters, each being equal either L, or R, or U, or D. k-th letter stands for the direction which Robot traveled the k-th unit segment in: L means that it moved to the left, R — to the right, U — to the top and D — to the bottom. Have a look at the illustrations for better explanation.

Output

The only line of input should contain the minimum possible length of the sequence.

Examples

input

4RURD

output

2

input

6RRULDD

output

2

input

26RRRULURURUULULLLDLDDRDRDLD

output

7

input

3RLL

output

2

input

4LRLR

output

4

Note

The illustrations to the first three tests are given below.

The last example illustrates that each point in the sequence should be counted as many times as it is presented in the sequence.

Source

Codeforces Round #389 (Div. 2, Rated, Based on Technocup 2017 - Elimination Round 3)

My Solution

题意：给定一个路径，让找出尽可能少的关键点，没相邻的2个关键点的路径必须是最短路径

贪心+最短路

这是一个很有趣的题 Y ^_^ Y

ans = 0;

(px,py)表示最近出现的一个关键点坐标，cnt表示从最近的那个关键点到当前点所走的路程， (x, y)表示当前所在的点的坐标。 

然后每步更新cnt及(x, y)后 判断abs(px - x) + abs(py - y) == cnt，如果成立则从最近的一个关键点到当前点一直在根据最短路走，直到每次出现不合理的情况，则上一个点更新为新的最近的关键点，并且ans++，扫一遍，最后一个点必然是关键点，故额外 ans++ 把最后一个点加上。

复杂度 O(n)

#include <iostream>#include <cstdio>#include <string>#include <cmath>using namespace std;typedef long long LL;const int maxn = 1e6 + 8;string s;int main(){    #ifdef LOCAL    freopen("c.txt", "r", stdin);    //freopen("c.out", "w", stdout);    int T = 8;    while(T--){    #endif // LOCAL    ios::sync_with_stdio(false); cin.tie(0);    int n, px = 0, py = 0, x = 0, y = 0, rx = 0, ry = 0, ans = 0, cnt = 0;    cin >> n;    cin >> s;    bool flag = false;    for(int i = 0; i < n; i++){        rx = x, ry = y;        if(s[i] == 'L'){            x--;        }        else if(s[i] == 'R'){            x++;        }        else if(s[i] == 'U'){            y++;        }        else if(s[i] == 'D'){            y--;        }        cnt++;        if(abs(px - x) + abs(py - y) == cnt){            //if(!flag){ans++; flag = true; }        }        else{            ans++;            //if(flag) ans++;            cnt = 1;            px = rx;            py = ry;            //cout << px << ' ' << py << endl;        }    }    ans++;    cout << ans << endl;    #ifdef LOCAL    cout << endl;    }    #endif // LOCAL    return 0;}

  Thank you!

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