Feed the dogs
| Time Limit: 6000MS | Memory Limit: 65536K | |
| Total Submissions: 7153 | Accepted: 1910 |
Description
Wind loves pretty dogs very much, and she has n pet dogs. So Jiajia has to feed the dogs every day for Wind. Jiajia loves Wind, but not the dogs, so Jiajia use a special way to feed the dogs. At lunchtime, the dogs will stand on one line, numbered from 1 to n, the leftmost one is 1, the second one is 2, and so on. In each feeding, Jiajia choose an inteval[i,j], select the k-th pretty dog to feed. Of course Jiajia has his own way of deciding the pretty value of each dog. It should be noted that Jiajia do not want to feed any position too much, because it may cause some death of dogs. If so, Wind will be angry and the aftereffect will be serious. Hence any feeding inteval will not contain another completely, though the intervals may intersect with each other.
Your task is to help Jiajia calculate which dog ate the food after each feeding.
Your task is to help Jiajia calculate which dog ate the food after each feeding.
Input
The first line contains n and m, indicates the number of dogs and the number of feedings.
The second line contains n integers, describe the pretty value of each dog from left to right. You should notice that the dog with lower pretty value is prettier.
Each of following m lines contain three integer i,j,k, it means that Jiajia feed the k-th pretty dog in this feeding.
You can assume that n<100001 and m<50001.
The second line contains n integers, describe the pretty value of each dog from left to right. You should notice that the dog with lower pretty value is prettier.
Each of following m lines contain three integer i,j,k, it means that Jiajia feed the k-th pretty dog in this feeding.
You can assume that n<100001 and m<50001.
Output
Output file has m lines. The i-th line should contain the pretty value of the dog who got the food in the i-th feeding.
Sample Input
7 2 1 5 2 6 3 7 4 1 5 3 2 7 1
Sample Output
3 2
题目大意：查询区间第K小数
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 using namespace std;
6 const int N = 100011;
7
8 int a[N];
9 int b[N];
10 int ans;
11 int res[50011];
12 struct node
13 {
14 int l;
15 int r;
16 int mid;
17 int cnt;
18 };
19 struct seg
20 {
21 int s,t;
22 int k;
23 int index;
24 };
25 bool cmp(seg a,seg b)
26 {
27 if(a.s==b.s)return a.t < b.t;
28 return a.s < b.s;
29 }
30 node seg_tree[3*N];
31 seg segs[50011];
32
33 void make(int l,int r,int num)
34 {
35 seg_tree[num].l = l;
36 seg_tree[num].r = r;
37 seg_tree[num].mid = (l+r)/2;
38 seg_tree[num].cnt = 0;
39 if(l+1!=r)
40 {
41 make(l,seg_tree[num].mid,2*num);
42 make(seg_tree[num].mid,r,2*num+1);
43 }
44 }
45 void insert(int val,int num)
46 {
47 if(seg_tree[num].l+1==seg_tree[num].r)
48 {
49 seg_tree[num].cnt++;
50 return;
51 }
52 if(val<seg_tree[num].mid)
53 {
54 insert(val,2*num);
55 }
56 else
57 {
58 insert(val,2*num+1);
59 }
60 seg_tree[num].cnt++;
61 }
62 void del(int val,int num)
63 {
64 if(seg_tree[num].l+1==seg_tree[num].r)
65 {
66 seg_tree[num].cnt--;
67 return;
68 }
69 if(val<seg_tree[num].mid)
70 {
71 del(val,2*num);
72 }
73 else
74 {
75 del(val,2*num+1);
76 }
77 seg_tree[num].cnt--;
78 }
79 void cal(int k,int num)
80 {
81 if(seg_tree[num].l+1==seg_tree[num].r)
82 {
83 ans = a[seg_tree[num].l];
84 return;
85 }
86 if(k<=seg_tree[2*num].cnt)
87 {
88 cal(k,2*num);
89 }
90 else
91 {
92 cal(k-seg_tree[2*num].cnt,2*num+1);
93 }
94 }
95
96 int n,m;
97
98 int b_search(int v)
99 {
100 int l = 1,r = n+1;
101 while(l<r)
102 {
103 int m = l + (r-l)/2;
104 if(a[m]==v)return m;
105 else if(a[m]>v)
106 {
107 r = m;
108 }
109 else
110 {
111 l = m+1;
112 }
113 }
114 return -1;
115 }
116
117 int main()
118 {
119 scanf("%d%d",&n,&m);
120 make(1,n+1,1);
121 for(int i=1;i<=n;i++)
122 {
123 scanf("%d",a+i);
124 b[i] = a[i];
125 }
126 sort(a+1,a+n+1);
127
128 int st,ed,k;
129 for(int i=0;i<m;i++)
130 {
131 scanf("%d%d%d",&st,&ed,&k);
132 if(st>ed)swap(st,ed);
133 segs[i].s = st;
134 segs[i].t = ed;
135 segs[i].k = k;
136 segs[i].index = i;
137 }
138 sort(segs,segs+m,cmp);
139 int prest = 0,preed =0;
140 for(int i=0;i<m;i++)
141 {
142 for(int j=prest;j<=min(segs[i].s-1,preed);j++)
143 {
144 if(j!=0)
145 {
146 del(b_search(b[j]),1);
147 }
148 }
149 for(int j=max(preed+1,segs[i].s);j<=segs[i].t;j++)
150 {
151 insert(b_search(b[j]),1);
152 }
153 cal(segs[i].k,1);
154 res[segs[i].index] = ans;
155 prest = segs[i].s;
156 preed = segs[i].t;
157
158 }
159 for(int i=0;i<m;i++)
160 {
161 printf("%d\n",res[i]);
162 }
163 return 0;
164 }
165
