outputstandard output
Skier rides on a snowy field. Its movements can be described by a string of characters ‘S’, ‘N’, ‘W’, ‘E’ (which correspond to 1 meter movement in the south, north, west or east direction respectively).

It is known that if he moves along a previously unvisited segment of a path (i.e. this segment of the path is visited the first time), then the time of such movement is 5 seconds. If he rolls along previously visited segment of a path (i.e., this segment of the path has been covered by his path before), then it takes 1 second.

Find the skier’s time to roll all the path.

Input
The first line contains an integer t (1≤t≤104) — the number of test cases in the input. Then t test cases follow.

Each set is given by one nonempty string of the characters ‘S’, ‘N’, ‘W’, ‘E’. The length of the string does not exceed 105 characters.

The sum of the lengths of t given lines over all test cases in the input does not exceed 105.

Output
For each test case, print the desired path time in seconds.

Example
inputCopy
5
NNN
NS
WWEN
WWEE
NWNWS
outputCopy
15
6
16
12
25
思路：我們判斷這條邊有沒有走過，主要就是看這條邊的兩個點有沒有同時出現過，同時出現過就是走過，否則就沒有。
主要用到了map。
代碼如下：

#include<bits/stdc++.h> #define ll long long using namespace std;string s; map<pair<int,int>,int> mp,mp1; int main() {int t;scanf("%d",&t);while(t--){mp.clear();mp1.clear();cin>>s;int ans=0;int cnt=0;mp[make_pair(0,0)]=++cnt;int x=0,y=0;int pre;for(int i=0;i<s.length();i++){pre=mp[make_pair(x,y)];if(s[i]=='N') x--;else if(s[i]=='S') x++;else if(s[i]=='W') y--;else y++;if(mp[make_pair(x,y)]==0) mp[make_pair(x,y)]=++cnt;int now=mp[make_pair(x,y)];if(mp1[make_pair(pre,now)]==0&&mp1[make_pair(now,pre)]==0) ans+=5;else ans+=1;mp1[make_pair(pre,now)]=mp1[make_pair(now,pre)]=1;}cout<<ans<<endl;}return 0; }

努力加油a啊，(^o)/~

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