Problem Description
Chinese always have the railway tickets problem because of its' huge amount of passangers and stations. Now goverment need you to develop a new tickets query system.
One train can just take k passangers. And each passanger can just buy one ticket from station a to station b. Each train cannot take more passangers any time. The one who buy the ticket earlier which can be sold will always get the ticket.
One train can just take k passangers. And each passanger can just buy one ticket from station a to station b. Each train cannot take more passangers any time. The one who buy the ticket earlier which can be sold will always get the ticket.
Input
The input contains servel test cases. The first line is the case number. In each test case:
The first line contains just one number k( 1 ≤ k ≤ 1000 ) and Q( 1 ≤ Q ≤ 100000 )
The following lines, each line contains two integers a and b, ( 1 ≤ a < b ≤ 1000000 ), indicate a query.
Huge Input, scanf recommanded.
The first line contains just one number k( 1 ≤ k ≤ 1000 ) and Q( 1 ≤ Q ≤ 100000 )
The following lines, each line contains two integers a and b, ( 1 ≤ a < b ≤ 1000000 ), indicate a query.
Huge Input, scanf recommanded.
Output
For each test case, output three lines:
Output the case number in the first line.
If the ith query can be satisfied, output i. i starting from 1. output an blank-space after each number.
Output a blank line after each test case.
Output the case number in the first line.
If the ith query can be satisfied, output i. i starting from 1. output an blank-space after each number.
Output a blank line after each test case.
题意：告诉你一辆车最多可以乘k个人，一共有q个人想要上车。按顺序给出每个人的开始的车站和结束的车站，问你有几个人能上的了车。
这题就是简单的区间修改，先定义所有区间为k，再求一下区间的最小值，如果最小值小于等于0那么就载不下更多人。每次更新区间都将
这个区间的所有值减1，然后更新一下最小值。最后要注意一下的是一个人上下车的时间是左闭右开区间，可以想到要是一个人下车，另一个人上车，这个情况下这个
点的大小还是不变的。
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
typedef long long ll;
const int M = 1e6 + 10;
int a[M] , b[M] , c[M];
struct TnT {
int l , r , min , add;
ll sum;
}T[M << 2];
int k , q;
void push(int p) {
if(T[p].add) {
T[p << 1].sum -= T[p].add * (T[p].r - T[p].l + 1);
T[p << 1].min -= T[p].add;
T[(p << 1) | 1].sum -= T[p].add * (T[p].r - T[p].l + 1);
T[(p << 1) | 1].min -= T[p].add;
T[p << 1].add += T[p].add;
T[(p << 1) | 1].add += T[p].add;
T[p].add = 0;
}
}
void build(int l , int r , int p) {
int mid = (l + r) >> 1;
T[p].l = l , T[p].r = r , T[p].add = 0;
if(T[p].l == T[p].r) {
T[p].sum = k;
T[p].min = k;
return ;
}
build(l , mid , p << 1);
build(mid + 1 , r , (p << 1) | 1);
T[p].sum = T[p << 1].sum + T[(p << 1) | 1].sum;
T[p].min = min(T[p << 1].min , T[(p << 1) | 1].min);
}
void updata(int l , int r , int p , int c) {
int mid = (T[p].l + T[p].r) >> 1;
if(T[p].l == l && T[p].r == r) {
T[p].sum -= c * (r - l + 1);
T[p].add += c;
T[p].min -= c;
return ;
}
push(p);
if(mid < l) {
updata(l , r , (p << 1) | 1 , c);
}
else if(mid >= r) {
updata(l , r , p << 1 , c);
}
else {
updata(l , mid , p << 1 , c);
updata(mid + 1 , r , (p << 1) | 1 , c);
}
T[p].sum = T[p << 1].sum + T[(p << 1) | 1].sum;
T[p].min = min(T[p << 1].min , T[(p << 1) | 1].min);
}
int query(int l , int r , int p) {
int mid = (T[p].l + T[p].r) >> 1;
if(T[p].l == l && T[p].r == r) {
return T[p].min;
}
push(p);
if(mid < l) {
return query(l , r , (p << 1) | 1);
}
else if(mid >= r) {
return query(l , r , p << 1);
}
else {
return min(query(l , mid , p << 1) , query(mid + 1 , r , (p << 1) | 1));
}
}
int main()
{
int n;
scanf("%d" , &n);
int ans = 0;
while(n--) {
scanf("%d%d" , &k , &q);
ans++;
int MAX = 0;
for(int i = 1 ; i <= q ; i++) {
scanf("%d%d" , &a[i] , &b[i]);
b[i]--;
MAX = max(MAX , max(a[i] , b[i]));
}
build(1 , MAX , 1);
int count = 0;
for(int i = 1 ; i <= q ; i++) {
int re = query(a[i] , b[i] , 1);
if(re <= 0) {
continue;
}
else {
updata(a[i] , b[i] , 1 , 1);
c[count++] = i;
}
}
printf("Case %d:
" , ans);
for(int i = 0 ; i < count ; i++) {
printf("%d " , c[i]);
}
printf("
");
}
return 0;
}