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:
-
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.
-
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
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:
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxn=5e6+10; 5 #define lson l,m,rt<<1 6 #define rson m+1,r,rt<<1|1 7 8 struct Tree{ 9 int lch,rch,val,lazy; 10 }tree[maxn<<2]; 11 12 void pushup(int l,int r,int rt) 13 { 14 int m=(l+r)>>1; 15 if(tree[rt<<1].val==(m-l+1)) tree[rt].lch=tree[rt<<1].val+tree[rt<<1|1].lch; 16 else tree[rt].lch=tree[rt<<1].lch; 17 if(tree[rt<<1|1].val==(r-m)) tree[rt].rch=tree[rt<<1|1].val+tree[rt<<1].rch; 18 else tree[rt].rch=tree[rt<<1|1].rch; 19 tree[rt].val=max(max(tree[rt<<1].val,tree[rt<<1|1].val),tree[rt<<1].rch+tree[rt<<1|1].lch); 20 } 21 22 void pushdown(int l,int r,int rt) 23 { 24 int m=(l+r)>>1; 25 if(tree[rt].lazy){ 26 if(tree[rt].lazy==1){//清空 27 tree[rt<<1].lazy=tree[rt<<1|1].lazy=tree[rt].lazy; 28 tree[rt<<1].lch=tree[rt<<1].rch=tree[rt<<1].val=m-l+1; 29 tree[rt<<1|1].lch=tree[rt<<1|1].rch=tree[rt<<1|1].val=r-m; 30 } 31 if(tree[rt].lazy==2){//坐满 32 tree[rt<<1].lazy=tree[rt<<1|1].lazy=tree[rt].lazy; 33 tree[rt<<1].lch=tree[rt<<1].rch=tree[rt<<1].val=0; 34 tree[rt<<1|1].lch=tree[rt<<1|1].rch=tree[rt<<1|1].val=0; 35 } 36 tree[rt].lazy=0; 37 } 38 } 39 40 void build(int l,int r,int rt) 41 { 42 tree[rt].lazy=0; 43 if(l==r){ 44 tree[rt].lch=tree[rt].rch=tree[rt].val=1; 45 return ; 46 } 47 48 int m=(l+r)>>1; 49 build(lson); 50 build(rson); 51 pushup(l,r,rt); 52 } 53 54 void update(int L,int R,int c,int l,int r,int rt) 55 { 56 if(tree[rt].lazy){ 57 pushdown(l,r,rt); 58 } 59 60 if(L<=l&&r<=R){ 61 if(c==1){ 62 tree[rt].lch=tree[rt].rch=tree[rt].val=r-l+1; 63 } 64 if(c==2){ 65 tree[rt].lch=tree[rt].rch=tree[rt].val=0; 66 } 67 tree[rt].lazy=c; 68 return ; 69 } 70 71 int m=(l+r)>>1; 72 if(L<=m) update(L,R,c,lson); 73 if(R> m) update(L,R,c,rson); 74 pushup(l,r,rt); 75 } 76 77 int query(int c,int l,int r,int rt) 78 { 79 if(tree[rt].lazy){ 80 pushdown(l,r,rt); 81 } 82 83 if(l==r){ 84 return l; 85 } 86 87 int m=(l+r)>>1; 88 if(tree[rt<<1].val>=c) return query(c,lson); 89 else if(tree[rt<<1].rch+tree[rt<<1|1].lch>=c) return m-tree[rt<<1].rch+1; 90 else return query(c,rson); 91 } 92 93 int main() 94 { 95 int n,m; 96 scanf("%d%d",&n,&m); 97 build(1,n,1); 98 int num=0; 99 for(int i=1;i<=m;i++){ 100 char op[5]; 101 scanf("%s",op); 102 if(op[0]=='A'){ 103 int x; 104 scanf("%d",&x); 105 if(tree[1].val>=x){ 106 int ans=query(x,1,n,1); 107 update(ans,ans+x-1,2,1,n,1); 108 } 109 else num++; 110 } 111 else{ 112 int l,r; 113 scanf("%d%d",&l,&r); 114 update(l,r,1,1,n,1); 115 } 116 } 117 printf("%d\n",num); 118 }
