Title:
given n, q, (1 <= n <= 10 ^ 9, 1 <= q <= 10 ^ 6), n represents the number 1 ~ n are arranged in a row, then the q-th interrogation, interrogation 1 Action: set x inaccessible. 2 query operation: start of the first number of outputs can be accessed from x (including x)
Sample input
. 5. 3
. 1 2
2 2
2. 1
sample output
. 3
. 1
#include <cstdio>
#include <unordered_map>
using namespace std;
unordered_map<int , int> pre;
int find(int x){
if(!pre.count(x)) return x;
return pre[x] = find(pre[x]);
}
int main(){
int n, q;
int op, x, ans;
scanf("%d %d", &n, &q);
while(q--){
scanf("%d %d", &op, &x);
if(op == 1){
pre[x] = find(x+1);
}
else{
ans = find(x);
if(x > n)
printf("-1\n");
else
printf("%d\n", ans);
}
}
return 0;
}
Resolve
By disjoint-set, when an operation is set to the current x root root x + 1, so that in operation 2, the number of x by finding the roots can be accurately x onward can be accessed first, if is to themselves, if it is behind a number of other, also found a number of their own roots, which represents it must be accessible, not 1 is set in operation, you can quickly. N can be a great find because, not so much open space, so use the map to keep the number, instead of the adjacent type of presence, and to find more efficient unordered_map speed, so use unordered_map.
