Super Mario
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7265 Accepted Submission(s): 3127
Problem Description
Mario is world-famous plumber. His “burly” figure and amazing jumping ability reminded in our memory. Now the poor princess is in trouble again and Mario needs to save his lover. We regard the road to the boss’s castle as a line (the length is n), on every integer point i there is a brick on height hi. Now the question is how many bricks in [L, R] Mario can hit if the maximal height he can jump is H.
Input
The first line follows an integer T, the number of test data.
For each test data:
The first line contains two integers n, m (1 <= n <=10^5, 1 <= m <= 10^5), n is the length of the road, m is the number of queries.
Next line contains n integers, the height of each brick, the range is [0, 1000000000].
Next m lines, each line contains three integers L, R,H.( 0 <= L <= R < n 0 <= H <= 1000000000.)
For each test data:
The first line contains two integers n, m (1 <= n <=10^5, 1 <= m <= 10^5), n is the length of the road, m is the number of queries.
Next line contains n integers, the height of each brick, the range is [0, 1000000000].
Next m lines, each line contains three integers L, R,H.( 0 <= L <= R < n 0 <= H <= 1000000000.)
Output
For each case, output "Case X: " (X is the case number starting from 1) followed by m lines, each line contains an integer. The ith integer is the number of bricks Mario can hit for the ith query.
Sample Input
1
10 10
0 5 2 7 5 4 3 8 7 7
2 8 6
3 5 0
1 3 1
1 9 4
0 1 0
3 5 5
5 5 1
4 6 3
1 5 7
5 7 3
Sample Output
Case 1:
4
0
0
3
1
2
0
1
5
1
Source
题意：问[l，r]区间内小于等于h的数有多少，即询问区间内h为第几大。
思路：可持久化线段树。
1 //2017-09-07 2 #include <cstdio> 3 #include <cstring> 4 #include <iostream> 5 #include <algorithm> 6 #define ll long long 7 #define mid ((l+r)>>1) 8 9 using namespace std; 10 11 const int N = 100010; 12 const int M = N * 30; 13 struct node{ 14 int lson, rson, sum;//sum维护l到r的数有几个 15 }tree[M]; 16 //第i棵线段树为插入前i个数字所构成的权值线段树。 17 int root[N], arr[N], arr2[N], tot; 18 int n, m, q; 19 20 void init(){//将原数列排序并去重 21 tot = 0; 22 for(int i = 1; i <= n; i++) 23 arr2[i] = arr[i]; 24 sort(arr2+1, arr2+1+n); 25 m = unique(arr2+1, arr2+1+n)-arr2-1; 26 } 27 28 //离散化 29 int getID(int x){ 30 return lower_bound(arr2+1, arr2+1+m, x) - arr2; 31 } 32 33 int build(int l, int r){ 34 int id = ++tot; 35 tree[id].sum = 0; 36 if(l == r)return id; 37 if(l <= mid) 38 tree[id].lson = build(l, mid); 39 if(r > mid) 40 tree[id].rson = build(mid+1, r); 41 return id; 42 } 43 44 int update(int id, int pos, int value){ 45 int newroot = ++tot, tmp = newroot; 46 tree[newroot].sum = tree[id].sum + value; 47 int l = 1, r = m; 48 while(l < r){ 49 if(pos <= mid){ 50 tree[newroot].lson = ++tot; 51 tree[newroot].rson = tree[id].rson; 52 newroot = tree[newroot].lson; 53 id = tree[id].lson; 54 r = mid; 55 }else{ 56 tree[newroot].rson = ++tot; 57 tree[newroot].lson = tree[id].lson; 58 newroot = tree[newroot].rson; 59 id = tree[id].rson; 60 l = mid+1; 61 } 62 tree[newroot].sum = tree[id].sum + value; 63 } 64 return tmp; 65 } 66 67 int query(int ltree, int rtree, int k){ 68 int l = 1, r = m, ans = 0; 69 while(l < r){ 70 if(k <= mid){ 71 ltree = tree[ltree].lson; 72 rtree = tree[rtree].lson; 73 r = mid; 74 }else{ 75 ans += tree[tree[rtree].lson].sum - tree[tree[ltree].lson].sum; 76 ltree = tree[ltree].rson; 77 rtree = tree[rtree].rson; 78 l = mid+1; 79 } 80 } 81 if(l == r){ 82 if(k < l)return 0; 83 else return ans + tree[rtree].sum - tree[ltree].sum; 84 } 85 } 86 87 int main() 88 { 89 //freopen("dataN.txt", "r", stdin); 90 int T, kase = 0; 91 scanf("%d", &T); 92 while(T--){ 93 scanf("%d%d", &n, &q); 94 for(int i = 1; i <= n; i++) 95 scanf("%d", &arr[i]); 96 init(); 97 root[0] = build(1, m); 98 for(int i = 1; i <= n; i++){ 99 int pos = getID(arr[i]); 100 root[i] = update(root[i-1], pos, 1); 101 } 102 printf("Case %d: ", ++kase); 103 while(q--){ 104 int l, r, h; 105 scanf("%d%d%d", &l, &r, &h); 106 l++; r++; 107 int pos = getID(h);//找到第一个小于概数的位置 108 if(arr2[pos] > h)pos--; 109 printf("%d ", query(root[l-1], root[r], pos)); 110 } 111 } 112 113 return 0; 114 }