[ABC426C]Upgrade Required题解

Time Limit: 2 sec / Memory Limit: 1024 MiB

Score : 300 points

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Problem Statement

There are N versions of a certain OS, numbered 1,2,…,N in chronological order.
There are N PCs, and initially the OS version of the i-th PC is i.

Perform Q operations in order. The i-th operation is as follows:

• Upgrade all PCs whose current OS version is Xi​ or earlier to version Yi​(>Xi​). Then, print the number of PCs upgraded in this operation.

Note that for i<Q, the upgrades from the i-th operation are performed before proceeding to the (i+1)-th operation.

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Constraints

• All input values are integers.
• 2≤N≤106
• 1≤Q≤2×105
• 1≤Xi​<Yi​≤N

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# #约束

—所有输入值均为整数。

—— 2≤N≤106

—— 1≤Q≤2×105

—— 1≤Xi​<Yi​≤N

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Input

The input is given from Standard Input in the following format:

N Q
X1​ Y1​
X2​ Y2​
⋮
XQ​ YQ​

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# #输入

输入来自标准输入，格式如下：

N Q
X1​ Y1​
X2​ Y2​
⋮
XQ​ YQ​

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Output

Output Q lines.
The i-th line should contain the number of PCs upgraded in the i-th operation.

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# #输出

输出 Q 行。

i \第一行应该包含 i \第一次操作中升级的pc的数量。

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Sample Input 1

Copy

8 5
2 6
3 5
1 7
5 7
7 8

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Sample Output 1

Copy

2
1
0
3
7

This input contains five operations.

• Initially, the versions of the eight PCs are 1,2,3,4,5,6,7,8.
• In the first operation, PCs with version 2 or earlier are upgraded to version 6.
  • This operation upgrades two PCs, and the versions of the PCs become 6,6,3,4,5,6,7,8.
• In the second operation, PCs with version 3 or earlier are upgraded to version 5.
  • This operation upgrades one PC, and the versions of the PCs become 6,6,5,4,5,6,7,8.
• In the third operation, PCs with version 1 or earlier are upgraded to version 7.
  • This operation upgrades zero PCs, and the versions of the PCs become 6,6,5,4,5,6,7,8.
• In the fourth operation, PCs with version 5 or earlier are upgraded to version 7.
  • This operation upgrades three PCs, and the versions of the PCs become 6,6,7,7,7,6,7,8.
• In the fifth operation, PCs with version 7 or earlier are upgraded to version 8.
  • This operation upgrades seven PCs, and the versions of the PCs become 8,8,8,8,8,8,8,8.

有道 翻译

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###输出示例

2
1
0
3
7

这个输入包含五个操作。

—最初，8台pc的版本号为 1,2,3,4,5,6,7,8 。

—第一次操作中，将版本为 2 及之前版本的pc升级到版本为 6 。

—该操作升级两台pc，两台pc的版本号变为 6,6,3,4,5,6,7,8 。

—第二次操作中，将版本为 3 及之前版本的pc升级为版本为 5 的pc。

—该操作升级了一台PC， PC的版本号变为 6,6,5,4,5,6,7,8 。

—第三次操作中，将版本为 1 及之前的pc升级到版本为 7 。

—此操作升级0台pc， pc的版本号变为 6,6,5,4,5,6,7,8 。

—第四次操作将版本为 5 及之前版本的pc升级到版本为 7 。

—升级三台pc，升级后的pc版本号为 6,6,7,7,7,6,7,8 。

—第五次操作中，将版本为 7 及之前版本的pc升级到版本为 8 。

—本次操作升级7台pc， pc的版本号为 8,8,8,8,8,8,8,8 。

思路

用线段树记录即可。

代码见下

#include<bits/stdc++.h>
using namespace std;
long long n,q,x[1000006],y[1000006],op[1000006],dd=0;
long long te[4000006],lz[4000006],xx,siz[4000006];
void alz(int a1,long long v){
	te[a1]=siz[a1]*v;
	lz[a1]=v;
}//a1打标记值v
void dow(int a1){
	if(lz[a1]!=-1){
		alz(a1*2,lz[a1]);
		alz(a1*2+1,lz[a1]);		
	}
	lz[a1]=-1;
}//a1下发标记至儿子
void bu(int a1,int l,int r){
	siz[a1]=r-l+1;
	lz[a1]=-1;
	if(l==r){
		te[a1]=1;
		return ;
	}
	int mid=(l+r)/2;
	bu(a1*2,l,mid);
	bu(a1*2+1,mid+1,r);
	te[a1]=te[a1*2]+te[a1*2+1];
	return ;
}//l~r建树节点a1
void ci(int a1,int l,int r,int x,int y,long long v){
	if(l>=x&&r<=y){
		alz(a1,v);
		return ;
	}
	int mid=(l+r)/2;
	dow(a1);
	if(x<=mid){
		ci(a1*2,l,mid,x,y,v);
	}
	if(mid+1<=y){
		ci(a1*2+1,mid+1,r,x,y,v);
	}
	te[a1]=te[a1*2]+te[a1*2+1];
	return ;
}//x~y区间加v至l~r区间a1点
long long co(int a1,int l,int r,int x,int y){
	if(x<=l&&r<=y){
		return te[a1];
	}
	int mid=(l+r)/2;
	long long dbdb=0;
	dow(a1);
	if(mid>=x){
		dbdb+=co(a1*2,l,mid,x,y);
	}
	if(mid+1<=y){
		dbdb+=co(a1*2+1,mid+1,r,x,y);
	}
	return dbdb;
}//x~y区间问至l~r区间a1点	
int main(){
	cin>>n>>q;
	bu(1,1,n);
	for(int i=1;i<=q;i++){
		cin>>x[i]>>y[i];
		cout<<co(1,1,n,1,x[i])<<endl;
		ci(1,1,n,y[i],y[i],co(1,1,n,1,x[i])+co(1,1,n,y[i],y[i]));
		ci(1,1,n,1,x[i],0);
	}
	return 0;
}