写的时候思路想到了,但是不怎么会维护。 这儿贴一个比较好理解的维护方式,用的双端队列。
#include<bits/stdc++.h>
using namespace std;
#define int long long
#pragma GCC optimize(3)
typedef pair<int, int>PII;
#define pb push_back
const int N = 1e5 + 10;
PII a[N];
vector<int>v[N];
void cf(){
int n, q;
cin >> n >> q;
for (int i = 1; i <= n; i++)v[i].clear();
int ma = 0;
deque<PII>q2;
for (int i = 1; i <= n; i++){
cin >> a[i].first;
a[i].second = i;
q2.push_back(a[i]);
ma = max(ma, a[i].first);
}
int ti = 0;//回合数
while (1){
if (q2.front().first == ma)break;
ti++;
PII x = q2.front();
q2.pop_front();
PII y = q2.front();
q2.pop_front();
if (x.first>y.first){
q2.push_front(x);
q2.push_back(y);
v[x.second].pb(ti);
}
else if (x.first<y.first){
q2.push_back(x);
q2.push_front(y);
v[y.second].pb(ti);
}
}
int id = q2.front().second;//最大元素的id
while (q--){
int x, y;
cin >> x >> y;
if (x == id){
if (ti){//第二到第一的过程也胜利了一次
if (y<ti)cout << 0 << endl;
else cout << y - ti + 1 << endl;
}
else{//始终在第一位
cout << y << endl;
}
}
else if (v[x].empty()){
cout << 0 << endl;
}
else{
int idx = upper_bound(v[x].begin(), v[x].end(), y) - v[x].begin();
cout << idx << endl;
}
}
}
signed main(){
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int _ = 1;
cin >> _;
while (_--){
cf();
}
return 0;
}
标签:q2,int,Tournament,Fighting,push,ti,front,814,first
From: https://www.cnblogs.com/xjtfate/p/16610024.html