Codeforces 1296C（map）

map开二维

 

题意：t（1e3）组输入，n和长度为n（2e5）的字符串，两端删除任意长度使得剩下的字符串的效果为0.有答案则输出剩余的字符串的左右边界下标，否则输出0.

思路：所有n的和小于2e5，复杂度n。开一个二维数组来存，n范围限制，所以用map。

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <vector>
//const int maxn = 1e5+5;
#define ll long long
#define inf  0x3f3f3f3f
#define FOR(i,a,b) for( int i = a;i <= b;++i)
#define bug cout<<"--------------"<<endl
 
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
 
using namespace std;
const int maxn =2e5+5;
map<int,int>m[maxn<<1];
char c[210000];
int main()
{
    int q;
    scanf("%d",&q);
    while(q--)    
    {
        int n,x,y,ansl=0,ansr=0,minn=inf;
        cin>>n;
        cin>>c+1;
        for(int i=2*n+1;i>=1;i--)
            m[i].clear();
        x=n+1,y=n+1;
        m[x][y]=1;
        for(int i = 1;i <= n; ++i)
        {
            if(c[i] == 'L') x--;
            else if(c[i] =='R')x++;
            else if(c[i] == 'U') y++;
            else y--;
            if(m[x][y] != 0)
            {
                int temp = m[x][y];
                if(i-temp+1 < minn)
                {
                    minn = i-temp+1;
                    ansl = temp;
                    ansr = i;
                }
            }
            m[x][y] = i+1;
        }
        if(minn == inf){
            cout<<-1<<endl;
        }
        else 
        {
            cout<<ansl<<" "<<ansr<<endl;
        }
    }
}