中文题面
思路：维和两个区间 一个是女神区间 一个是基友区间 如果是基友要预约时间 直接在基友区间查询可满足的起点 （这里先判tree[1].m >=length也就是有没有这样的区间满足时间length） 预约成功后更新基友区间
如果是女神要预约区间 先在基友区间预约看有没有满足的区间 （同样看根节点的m） 如果有 同时更新两个区间 如果没有继续在女神区间查找 如果女神区间有 则同时更新基友和女神区间 如果没有 那就真没有了直接输出 题目要求的话
1 #include<bits/stdc++.h>
2 using namespace std;
3 const int maxn=1e6+5;
4 struct Node{
5 int l,r;
6 int ml,mr,m;
7 }tree1[maxn*4],tree2[maxn*4];
8 void build(int x,int l,int r){
9 tree1[x].l=tree2[x].l=l;
10 tree1[x].r=tree2[x].r=r;
11 tree1[x].ml=tree2[x].ml=tree1[x].mr=tree2[x].mr=tree1[x].m=tree2[x].m=r-l+1;
12 if(l==r)return ;
13 int mid=l+r>>1;
14 build(x<<1,l,mid);
15 build(x<<1|1,mid+1,r);
16 }
17 void push_up(int x){
18 int mid=tree1[x].r+tree1[x].l>>1;
19 tree1[x].ml=tree1[x<<1].ml;
20 tree1[x].mr=tree1[x<<1|1].mr;
21 tree1[x].m=max(max(tree1[x<<1].m,tree1[x<<1|1].m),tree1[x<<1].mr+tree1[x<<1|1].ml);
22 if(tree1[x<<1].ml==mid-tree1[x].l+1)tree1[x].ml+=tree1[x<<1|1].ml;
23 if(tree1[x<<1|1].mr==tree1[x].r-mid)tree1[x].mr+=tree1[x<<1].mr;
24
25 tree2[x].ml=tree2[x<<1].ml;
26 tree2[x].mr=tree2[x<<1|1].mr;
27 tree2[x].m=max(max(tree2[x<<1].m,tree2[x<<1|1].m),tree2[x<<1].mr+tree2[x<<1|1].ml);
28 if(tree2[x<<1].ml==mid-tree2[x].l+1)tree2[x].ml+=tree2[x<<1|1].ml;
29 if(tree2[x<<1|1].mr==tree2[x].r-mid)tree2[x].mr+=tree2[x<<1].mr;
30
31 }
32 void push_down(int x){
33 int mid=tree1[x].l+tree1[x].r>>1;
34 if(tree1[x].m==tree1[x].r-tree1[x].l+1){
35 tree1[x<<1].ml=tree1[x<<1].mr=tree1[x<<1].m=mid-tree1[x].l+1;
36 tree1[x<<1|1].ml=tree1[x<<1|1].mr=tree1[x<<1|1].m=tree1[x].r-mid;
37 }
38 else if(tree1[x].m==0){
39 tree1[x<<1].ml=tree1[x<<1].mr=tree1[x<<1].m=0;
40 tree1[x<<1|1].ml=tree1[x<<1|1].mr=tree1[x<<1|1].m=0;
41 }
42
43 if(tree2[x].m==tree2[x].r-tree2[x].l+1){
44 tree2[x<<1].ml=tree2[x<<1].mr=tree2[x<<1].m=mid-tree2[x].l+1;
45 tree2[x<<1|1].mr=tree2[x<<1|1].ml=tree2[x<<1|1].m=tree2[x].r-mid;
46 }
47 else if(tree2[x].m==0){
48 tree2[x<<1].ml=tree2[x<<1].mr=tree2[x<<1].m=0;
49 tree2[x<<1|1].mr=tree2[x<<1|1].ml=tree2[x<<1|1].m=0;
50 }
51 }
52 void update(int x,int l,int r,int value){
53 if(tree1[x].l>=l&&tree1[x].r<=r){
54 if(value==1){
55 tree1[x].ml=tree1[x].mr=tree1[x].m=0;
56 }
57 else if(value==2){
58 tree1[x].ml=tree1[x].mr=tree1[x].m=0;
59 tree2[x].ml=tree2[x].mr=tree2[x].m=0;
60 }
61 else {
62 tree1[x].ml=tree1[x].mr=tree1[x].m=tree1[x].r-tree1[x].l+1;
63 tree2[x].ml=tree2[x].mr=tree2[x].m=tree1[x].r-tree1[x].l+1;
64 }
65 return ;
66 }
67 int mid=tree1[x].l+tree1[x].r>>1;
68 push_down(x);
69 if(mid>=l)update(x<<1,l,r,value);
70 if(mid<r)update(x<<1|1,l,r,value);
71 push_up(x);
72 }
73 int query(int x,int length,int value){
74 if(tree1[x].l==tree1[x].r)return tree1[x].l;
75 push_down(x);
76 int mid=tree1[x].l+tree1[x].r>>1;
77 if(value==1){
78 if(tree1[x<<1].m>=length){
79 return query(x<<1,length,value);
80 }
81 else if(tree1[x<<1].mr+tree1[x<<1|1].ml>=length){
82 return mid-tree1[x<<1].mr+1;
83 }
84 else return query(x<<1|1,length,value);
85 }
86 else {
87 if(tree2[x<<1].m>=length){
88 return query(x<<1,length,value);
89 }
90 else if(tree2[x<<1].mr+tree2[x<<1|1].ml>=length){
91 return mid-tree2[x<<1].mr+1;
92 }
93 else return query(x<<1|1,length,value);
94 }
95 }
96 int main(){
97 int t;
98 int kase=1;
99 scanf("%d",&t);
100 char s[100];
101 int temp;
102 int q;
103 int n;
104 while(t--){
105 printf("Case %d:\n",kase++);
106 scanf("%d%d",&n,&q);
107 build(1,1,n);
108 while(q--){
109 scanf("%s%d",s,&temp);
110 if(s[0]==‘D‘){
111 if(tree1[1].m>=temp){
112 int start=query(1,temp,1);
113 update(1,start,start+temp-1,1);
114 printf("%d,let‘s fly\n",start);
115 }
116 else printf("fly with yourself\n");
117 }
118 else if(s[0]==‘N‘){
119 if(tree1[1].m>=temp){
120 int start=query(1,temp,1);
121 update(1,start,start+temp-1,2);
122 printf("%d,don‘t put my gezi\n",start);
123 }
124 else if(tree2[1].m>=temp){
125 int start=query(1,temp,0);
126 update(1,start,start+temp-1,2);
127 printf("%d,don‘t put my gezi\n",start);
128 }
129 else printf("wait for me\n");
130 }
131 else {
132 int zz;
133 scanf("%d",&zz);
134 update(1,temp,zz,3);
135 printf("I am the hope of chinese chengxuyuan!!\n");
136 }
137 }
138
139 }
140 return 0;
141 }
原文地址：https://www.cnblogs.com/ttttttttrx/p/10349570.html
时间： 2024-08-03 07:12:39