CF Round 881 --E. Tracking Segments
You are given an array a consisting of n zeros. You are also given a set of m not necessarily different segments. Each segment is defined by two numbers li and ri (1≤li≤ri≤n) and represents a subarray ali,ali+1,…,ar of the array a.
Let's call the segment li,ri beautiful if the number of ones on this segment is strictly greater than the number of zeros. For example, if a=[1,0,1,0,1], then the segment [1,5] is beautiful (the number of ones is 3, the number of zeros is 2), but the segment [3,4] is not is beautiful (the number of ones is 1, the number of zeros is 1).
You also have q changes. For each change you are given the number 1≤x≤n, which means that you must assign an element ax the value 1.
You have to find the first change after which at least one of m given segments becomes beautiful, or report that none of them is beautiful after processing all q changes.
Input
The first line contains a single integer t (1≤t≤10^4) — the number of test cases.
The first line of each test case contains two integers n and m (1≤m≤n≤10^5) — the size of the array a and the number of segments, respectively.
Then there are m lines consisting of two numbers li and ri (1≤li≤ri≤n) —the boundaries of the segments.
The next line contains an integer q (1≤q≤n) — the number of changes.
The following q lines each contain a single integer x (1≤x≤n) — the index of the array element that needs to be set to 1. It is guaranteed that indexes in queries are distinct.
It is guaranteed that the sum of n for all test cases does not exceed 10^5.
Output
For each test case, output one integer — the minimum change number after which at least one of the segments will be beautiful, or −1 if none of the segments will be beautiful.
题意：给定长度为n的数组，初始全是0，以及m个子段区间，然后q次修改，每次修改会使a[x]变成1，问你最少前k次修改可以使得至少一个子段区间是美丽的或者q次都无法满足输出-1，区间美丽需要满足区间内1的个数严格大于0的个数。
解析：很明显就是单点修改，区间查询加二分答案，我们可以直接用树状数组来维护区间和，如果某个区间和＞区间长度的一半，那么就说明1的个数严格＞0的个数，那么就是完美的。
#include
using namespace std;
const int N=1e5+5;
int x[N],y[N],pos[N];//记录子段的左右区间以及每个修改的位置
int tr[N];
int n,m,q;
int lowbit(int x)
{
return x&-x;
}
int add(int x,int c)
{
for(int i=x;i<=n;i+=lowbit(i)) tr[i]+=c;
}
int sum(int x)//求tr[1~x]的前缀和
{
int res=0;
for(int i=x;i;i-=lowbit(i)) res+=tr[i];
return res;
}
bool check(int k)
{
for(int i=1;i<=n;i++) tr[i]=0;//记得要初始化tr数组
for(int i=1;i<=k;i++) add(pos[i],1);//对应位置修改成1
for(int i=1;i<=m;i++)
{
int len=y[i]-x[i]+1;
if(sum(y[i])-sum(x[i]-1)>len/2) return true;//如果满足一个就返回true
}
return false;
}
void solve()
{
scanf("%d%d",&n,&m);
for(int i=1;i<=m;i++) scanf("%d%d",&x[i],&y[i]);
scanf("%d",&q);
for(int i=1;i<=q;i++) scanf("%d",&pos[i]);
int l=1,r=q,ans=-1;
while(l<=r)//二分答案
{
int mid=l+r>>1;
if(check(mid)) ans=mid,r=mid-1;
else l=mid+1;
}
printf("%d\n",ans);
}
int main()
{
int t=1;
scanf("%d",&t);
while(t--) solve();
return 0;
}