Command Sequence

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题意

一个机器人能上下左右移动。给定一个移动的字符串，问有多少的子串可以使其按照子串的顺序来走能回到原点

题解

我打眼望过去，这本质上就是两种类型的括号匹配，要两种括号要同时匹配成功

比赛的时候讨论好解法时，当发来一个片段的时候：

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到最后也没优化出来

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <map>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <sstream>
#include <unordered_map>
#define ll long long
#define ull unsigned long long
#define re return
#define pb push_back
#define Endl "\n"
#define endl "\n"
#define x first
#define y second

using namespace std;

typedef pair<int, int> PII;

const int N = 2e5 + 10;
const int M = 1e5 + 10;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;

int dx[4] = {-1,0,1,0};
int dy[4] = {0,1,0,-1};

struct nope{
    int id;
    int x;
} a[N];

int T;
int n;
char s[N];
int s1[N]; // ud
int s2[N]; // lr

int main(){
    scanf("%d", &T);
    while(T--){

        scanf("%d", &n);
        scanf("%s", s + 1);
        // vector<int> v1[N];
        // vector<int> v2[N];

        for (int i = 1; i <= n; i++){
            if(s[i] == 'U')
                a[i].x = 1, a[i].id = 1;
            if(s[i] == 'D')
                a[i].x = -1, a[i].id = 1;
            if(s[i] == 'L')
                a[i].x = 1, a[i].id = 2;
            if(s[i] == 'R')
                a[i].x = -1, a[i].id = 2;
        }

        // int maxx = 0;

        for (int i = 1; i <= n; i++){
            if(a[i].id == 1){ //UD
                s1[i] = s1[i - 1] + a[i].x;
                s2[i] = s2[i - 1];
            }
            else{ // LR
                s2[i] = s2[i - 1] + a[i].x;
                s1[i] = s1[i - 1];
            }
            // maxx = max(maxx, max(s1[i], s2[i]));
        }

        for (int i = 1; i <= n; i++){
            cout << s1[i] << " ";
        }
        cout << endl;
        for (int i = 1; i <= n; i++){
            cout << s2[i] << " ";
        }
        cout << endl;

        ll ans = 0;
        for (int i = 0; i < n; i++){ // 到最后也不知道怎么优化

        }

        printf("%lld\n", ans);
    }
}

赛后根据上面的那个公式，我们把代码做了一下优化

相信只要懂了上面的聊天记录就能看懂下面的代码

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <map>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <sstream>
#include <unordered_map>
#define ll long long
#define ull unsigned long long
#define re return
#define pb push_back
#define Endl "\n"
#define endl "\n"
#define x first
#define y second

using namespace std;

typedef pair<int, int> PII;

const int N = 2e5 + 10;
const int M = 1e5 + 10;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;

int dx[4] = {-1,0,1,0};
int dy[4] = {0,1,0,-1};

int T;
int n;
char s[N];
int a[N];

int main(){
    cin >> T;
    while(T--){
        scanf("%d", &n);
        scanf("%s", s + 1);

        for (int i = 1; i <= n; i++){
            if(s[i] == 'U')
                a[i] = 1;
            if(s[i] == 'D')
                a[i] = -1;
            if(s[i] == 'L')
                a[i] = 100000; // 设1e5是为了与数据范围相符合
            if(s[i] == 'R')
                a[i] = -100000;
        }

        map<ll, int> m;
        ll pre = 0;
        ll ans = 0;

        m[pre]++;

        for (int i = 1; i <= n; i++){
            pre += a[i];
            if(m.find(pre) != m.end())
                ans += m[pre];
            m[pre]++;
        }

        printf("%lld\n", ans);
    }
}