POJ 3277 City Horizon （离散化线段树）
链接：http://poj.org/problem?id=3277
Description
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 line with N (1 ≤ N ≤ 40,000) buildings. Building i's silhouette has a base that spans locations Ai through Bi along the horizon (1 ≤ Ai < Bi ≤ 1,000,000,000) and has height Hi(1 ≤ Hi ≤ 1,000,000,000). Determine the area, in square units, of the aggregate silhouette formed by all N buildings.
Input
Line 1: A single integer: N
Lines 2..N+1: Input line i+1 describes building i with three space-separated integers: Ai, Bi, and Hi
Lines 2..N+1: Input line i+1 describes building i with three space-separated integers: Ai, Bi, and Hi
Output
Line 1: The total area, in square units, of the silhouettes formed by all
N buildings
Sample Input
4
2 5 1
9 10 4
6 8 2
4 6 3
Sample Output
16
Hint
The first building overlaps with the fourth building for an area of 1 square unit, so the total area is just 3*1 + 1*4 + 2*2 + 2*3 - 1 = 16.
思路：
1>. 建立线段树，求出每一额线段的高度，进而求出面积；
2>. Ai 和 Bi的范围都很大，所以不能直接建立线段树，用离散化的方法建树；
代码如下：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#define RST(N)memset(N, 0, sizeof(N))
using namespace std;
typedef __int64 INT;
typedef struct Building_
{
INT x1, x2, height;
}Building;
struct Trie
{
INT left, right, hi;
}trie[1000000];
Building N[50000];
INT Node[100000]; //表示离散化后的端点个数；
bool cmp(Building a, Building b) { return a.height < b.height; }
void Init(int n) //初始化；
{
RST(N), RST(trie), RST(Node);
for(int i=1; i<=n; i++) {
scanf("%I64d %I64d %I64d", &N[i].x1, &N[i].x2, &N[i].height);
Node[i*2-1] = N[i].x1, Node[i*2] = N[i].x2;
}
sort(N+1, N+n+1, cmp); //端点排序；
sort(Node+1, Node+2*n+1);
int j = 1;
for(int i=1; i<=2*n; i++) { //离散化；
if(Node[i] != Node[j]) Node[++j] = Node[i];
}
Node[0] = j;
return ;
}
void build(INT L, INT R, INT p)
{
trie[p].left = Node[L];
trie[p].right = Node[R];
if(L + 1 == R) return ;
int mid = (L + R)/2;
build(L, mid, p*2);
build(mid, R, p*2+1);
return ;
}
void update(INT L, INT R, INT value, INT p)
{
if(L <= trie[p].left && trie[p].right <= R) { //当前线段被包含与插入的线段；
trie[p].hi = value;
return ;
}
if(trie[p].hi > 0) trie[p*2].hi = trie[p*2+1].hi = trie[p].hi;
trie[p].hi = -1;
if(R > trie[p*2].right) update(L, R, value, p*2+1);
if(L < trie[p*2].right) update(L, R, value, p*2);
return ;
}
INT Query(INT p)
{
if(trie[p].hi > 0) return trie[p].hi*(trie[p].right-trie[p].left);
if(trie[p].hi == 0) return 0;
return (Query(p*2) + Query(p*2+1));
}
int main()
{
int n;
scanf("%d", &n);
Init(n);
build(1, Node[0], 1);
for(int i=1; i<=n; i++) update(N[i].x1, N[i].x2, N[i].height, 1);
printf("%I64d\n", Query(1));
}
做这道题的时候，有一个问题我一直都没弄明白，，为什么用__int64能过，用unsigned long long不能过。。。求大牛解释；