HDU 6012【離散化+掃描線】

Lotus and Horticulture

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 698    Accepted Submission(s): 224

Problem Description

These days Lotus is interested in cultivating potted plants, so she wants to build a greenhouse to meet her research desires.

Lotus placed all of the n pots in the new greenhouse, so all potted plants were in the same environment.

Each plant has an optimal growth temperature range of [l,r], which grows best at this temperature range, but does not necessarily provide the best research value (Lotus thinks that researching poorly developed potted plants are also of great research value).

Lotus has carried out a number of experiments and found that if the growth temperature of the i-th plant is suitable, it can provide ai units of research value; if the growth temperature exceeds the upper limit of the suitable temperature, it can provide the bi units of research value; temperatures below the lower limit of the appropriate temperature, can provide ci units of research value.

Now, through experimentation, Lotus has known the appropriate growth temperature range for each plant, and the values of a, b, c are also known. You need to choose a temperature for the greenhouse based on these information, providing Lotus with the maximum research value.

__NOTICE: the temperature can be any real number.__

 

 

Input

The input includes multiple test cases. The first line contains a single integer T, the number of test cases.

The first line of each test case contains a single integer n∈[1,50000], the number of potted plants.

The next n line, each line contains five integers li,ri,ai,bi,ci∈[1,109].

 

 

Output

For each test case, print one line of one single integer presenting the answer.

 

 

Sample Input

1 5 5 8 16 20 12 10 16 3 13 13 8 11 13 1 11 7 9 6 17 5 2 11 20 8 5

 

 

Sample Output

83

 

題意：

n個區間 [)前閉後開型，每一個區間左中右分別三個花費，問哪一個點花費最大。

題解：

由於點能夠取任意實數，區間是整數，因此距離區間端點0.5的點都是可能的點，將區間左右端點*2，每一個區間離散化成 l-1, l, r這三個點。而後掃描線去遍歷，碰到左邊邊界 +a-c; 右邊界 +b-a;

代碼:

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstring>
 4 #include <cstdio>
 5 #include <bitset>
 6 #include <vector>
 7 #include <queue>
 8 #include <stack>
 9 #include <cmath>
10 #include <list>
11 #include <set>
12 #include <map>
13 #define rep(i,a,b) for(int i = a;i <= b;++ i)
14 #define per(i,a,b) for(int i = a;i >= b;-- i)
15 #define mem(a,b) memset((a),(b),sizeof((a)))
16 #define FIN freopen("in.txt","r",stdin)
17 #define FOUT freopen("out.txt","w",stdout)
18 #define IO ios_base::sync_with_stdio(0),cin.tie(0)
19 #define mid ((l+r)>>1)
20 #define ls (id<<1)
21 #define rs ((id<<1)|1)
22 #define N 50005
23 #define INF 0x3f3f3f3f
24 #define INFF ((1LL<<62)-1)
25 typedef long long LL;
26 using namespace std;
27 
28 int T, n, HASH[N*3], l[N], r[N], a[N], b[N], c[N];
29 LL cost[N*3];
30 int main()
31 {IO;
32     //FIN;
33     cin >> T;
34     while(T--){
35         cin >> n;
36         int len = 0;
37         LL tol = 0;
38         rep(i, 1, n){
39             cin >> l[i] >> r[i] >> a[i] >> b[i] >> c[i];
40             l[i] <<= 1; r[i] <<= 1;
41             HASH[len++] = l[i]; HASH[len++] = l[i]-1; HASH[len++] = r[i];
42             tol += c[i];
43         }
44         sort(HASH, HASH+len);
45         int size = unique(HASH, HASH+len)-HASH;
46         LL ans = tol;
47         mem(cost, 0);
48         
49         rep(i, 1, n){
50             l[i] = lower_bound(HASH, HASH+size, l[i])-HASH;
51             r[i] = lower_bound(HASH, HASH+size, r[i])-HASH;
52             cost[l[i]] += a[i]-c[i];
53             cost[r[i]+1] += b[i]-a[i];
54         }
55         rep(i, 0, size-1){
56             tol += cost[i];
57             ans = max(ans, tol);
58         }
59         cout << ans << endl;
60     }
61     return 0;
62 }

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