洛谷 P3071 [USACO13JAN]座位Seating-线段树区间合并(判断找，只需要最大前缀和最大后缀)+分治+贪心

P3071 [USACO13JAN]座位Seating

题目描述

To earn some extra money, the cows have opened a restaurant in their barn specializing in milkshakes. The restaurant has N seats (1 <= N <= 500,000) in a row. Initially, they are all empty.

Throughout the day, there are M different events that happen in sequence at the restaurant (1 <= M <= 300,000). The two types of events that can happen are:

1. A party of size p arrives (1 <= p <= N). Bessie wants to seat the party in a contiguous block of p empty seats. If this is possible, she does so in the lowest position possible in the list of seats. If it is impossible, the party is turned away.

2. A range [a,b] is given (1 <= a <= b <= N), and everybody in that range of seats leaves.

Please help Bessie count the total number of parties that are turned away over the course of the day.

有一排n个座位，m次操作。A操作：将a名客人安置到最左的连续a个空位中，没有则不操作。L操作：[a,b]的客人离开。

求A操作的失败次数。

输入输出格式

输入格式：

* Line 1: Two space-separated integers, N and M.

* Lines 2..M+1: Each line describes a single event. It is either a line of the form "A p" (meaning a party of size p arrives) or "L a b" (meaning that all cows in the range [a, b] leave).

输出格式：

* Line 1: The number of parties that are turned away.

输入输出样例

输入样例#1： 复制

10 4
A 6
L 2 4
A 5
A 2

输出样例#1： 复制

1

说明

There are 10 seats, and 4 events. First, a party of 6 cows arrives. Then all cows in seats 2..4 depart. Next, a party of 5 arrives, followed by a party of 2.

Party #3 is turned away. All other parties are seated.

代码:

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int maxn=5e6+;
 #define lson l,m,rt<<1
 #define rson m+1,r,rt<<1|1

 struct Tree{
     int lch,rch,val,lazy;
 }tree[maxn<<];

 void pushup(int l,int r,int rt)
 {
     int m=(l+r)>>;
     if(tree[rt<<].val==(m-l+)) tree[rt].lch=tree[rt<<].val+tree[rt<<|].lch;
     else tree[rt].lch=tree[rt<<].lch;
     if(tree[rt<<|].val==(r-m)) tree[rt].rch=tree[rt<<|].val+tree[rt<<].rch;
     else tree[rt].rch=tree[rt<<|].rch;
     tree[rt].val=max(max(tree[rt<<].val,tree[rt<<|].val),tree[rt<<].rch+tree[rt<<|].lch);
 }

 void pushdown(int l,int r,int rt)
 {
     int m=(l+r)>>;
     if(tree[rt].lazy){
         if(tree[rt].lazy==){//清空
             tree[rt<<].lazy=tree[rt<<|].lazy=tree[rt].lazy;
             tree[rt<<].lch=tree[rt<<].rch=tree[rt<<].val=m-l+;
             tree[rt<<|].lch=tree[rt<<|].rch=tree[rt<<|].val=r-m;
         }
         if(tree[rt].lazy==){//坐满
             tree[rt<<].lazy=tree[rt<<|].lazy=tree[rt].lazy;
             tree[rt<<].lch=tree[rt<<].rch=tree[rt<<].val=;
             tree[rt<<|].lch=tree[rt<<|].rch=tree[rt<<|].val=;
         }
         tree[rt].lazy=;
     }
 }

 void build(int l,int r,int rt)
 {
     tree[rt].lazy=;
     if(l==r){
         tree[rt].lch=tree[rt].rch=tree[rt].val=;
         return ;
     }

     int m=(l+r)>>;
     build(lson);
     build(rson);
     pushup(l,r,rt);
 }

 void update(int L,int R,int c,int l,int r,int rt)
 {
     if(tree[rt].lazy){
         pushdown(l,r,rt);
     }

     if(L<=l&&r<=R){
         if(c==){
             tree[rt].lch=tree[rt].rch=tree[rt].val=r-l+;
         }
         if(c==){
             tree[rt].lch=tree[rt].rch=tree[rt].val=;
         }
         tree[rt].lazy=c;
         return ;
     }

     int m=(l+r)>>;
     if(L<=m) update(L,R,c,lson);
     if(R> m) update(L,R,c,rson);
     pushup(l,r,rt);
 }

 int query(int c,int l,int r,int rt)
 {
     if(tree[rt].lazy){
         pushdown(l,r,rt);
     }

     if(l==r){
         return l;
     }

     int m=(l+r)>>;
     if(tree[rt<<].val>=c) return query(c,lson);
     else if(tree[rt<<].rch+tree[rt<<|].lch>=c) return m-tree[rt<<].rch+;
     else return query(c,rson);
 }

 int main()
 {
     int n,m;
     scanf("%d%d",&n,&m);
     build(,n,);
     int num=;
     for(int i=;i<=m;i++){
         char op[];
         scanf("%s",op);
         if(op[]=='A'){
             int x;
             scanf("%d",&x);
             if(tree[].val>=x){
                 int ans=query(x,,n,);
                 update(ans,ans+x-,,,n,);
             }
             else num++;
         }
         else{
             int l,r;
             scanf("%d%d",&l,&r);
             update(l,r,,,n,);
         }
     }
     printf("%d\n",num);
 }