Xtu data structure H. City horizon

H. City horizontime limit: 2000 msmemory limit: 65536kb64-bit integer Io format: % LLD Java class name: Main

Farmer John has taken his cows on a trip to the city! As the sun sets, the cows gaze at the city horizon and observe the beautiful silhouettes formed by the rectangular buildings.

The entire horizon is represented by a number lineN(1 ≤N≤ 40,000) buildings. BuildingI'S silhouette has a base that spans locationsAIThroughBiAlong the horizon (1 ≤AI<Bi≤ 1,000,000,000) and has heightHi(1 ≤Hi≤ 1,000,000,000). determine the area, in square units, of the aggregate silhouette formed by allNBuildings.

Inputline 1: A single INTEGER: N
Lines 2 .. N+ 1: input line I+ 1 describes Building IWith three space-separated integers: AI, Bi, And HiOutputline 1: The total area, in square units, of the silhouettes formed by all NBuildings sample input

42 5 19 10 46 8 24 6 3

Sample output

16

Problem solving: reading the report on the question is still learning about the gray machine of discretization. However, this part of the line segment tree is actually very easy, but with the discretization operation, eggache!

1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <cstdlib> 5 # include <vector> 6 # include <climits> 7 # include <algorithm> 8 # include <cmath> 9 # define ll long long10 # define INF 0x3f3f11 using namespace STD; 12 const int maxn = 40010; 13 struct building {14 LL a, B, H; 15} BB [maxn]; 16 struct node {17 int lt, RT, h; 18} tree [maxn <3]; 19 ll p [maxn <1], n, ans; 20 int CNT; 21 bool CMP (Co NST Building & X, const Building & Y) {22 return X. h <Y. h; 23} 24 void build (INT lt, int RT, int v) {25 tree [v]. lt = lt; 26 tree [v]. RT = RT; 27 tree [v]. H = 0; 28 If (RT-Lt = 1) return; 29 int mid = (LT + RT)> 1; 30 build (LT, mid, v <1); 31 build (MID, RT, v <1 | 1); 32} 33 void Update (INT lt, int RT, int Val, int V) {34 if (tree [v]. lt = lT & tree [v]. RT = RT) {35 Tree [v]. H = val; 36 return; 37} 38 If (tree [v]. h> 0) {39 tree [v <<1]. H = tree [v <1 | 1]. H = tree [v]. h; 40 tree [v]. H = 0; 41} 42 int mid = (tree [v]. lt + tree [v]. RT)> 1; 43 If (RT <= mid) Update (LT, RT, Val, v <1); 44 else if (LT> = mid) update (LT, RT, Val, v <1 | 1); 45 else {46 Update (LT, mid, Val, v <1); 47 Update (MID, RT, Val, v <1 | 1); 48} 49} 50 void query (INT v) {51 if (tree [v]. h> 0) {52 ans + = (LL) tree [v]. H * (p [tree [v]. RT-1]-P [tree [v]. lt-1]); 53 return; 54} 55 if (tree [v]. RT-tree [v]. lt = 1) return; 56 query (v <1); 57 query (v <1 | 1); 58} 59 int bsearch (INT lt, int RT, int Val) {60 while (LT <= RT) {61 int mid = (LT + RT)> 1; 62 if (P [Mid] = Val) 63 return mid + 1; 64 else if (Val <p [Mid]) 65 RT = mid-1; 66 else lT = Mid + 1; 67} 68 return 0; 69} 70 int main () {71 int I, j, R; 72 scanf ("% LLD", & N); 73 for (I = 0; I <N; I ++) {74 scanf ("% d", & BB [I]. a, & BB [I]. b, & BB [I]. h); 75 p [CNT ++] = BB [I]. a; 76 P [CNT ++] = B B [I]. B; 77} 78 sort (p, p + CNT); // The sorting is to use binary search. 79 sort (BB, BB + N, CMP); 80 build (1, n <1, 1); 81 r = (n <1)-1; 82 for (I = 0; I <n; I ++) {83 int lT = bsearch (0, cnt-1, BB [I]. a); 84 int RT = bsearch (0, cnt-1, BB [I]. b); 85 Update (LT, RT, BB [I]. h, 1); 86} 87 ans = 0; 88 query (1); 89 printf ("% LLD", ANS); 90 return 0; 91}

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