本来想打线段树的说。。。

就是把坐标离散化了,然后区间最大求和即可。。。

后来觉得有点烦的说(silver题就要线段树。。。),于是看了下usaco的题解,发现了个高端的东西:善用STL里的容器和迭代器就可以了。

以下就是高端程序:

 

 1 /**************************************************************
 2     Problem: 1645
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:264 ms
 7     Memory:6120 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11 #include <vector>
12 #include <utility>
13 #include <map>
14 #include <set>
15 #include <algorithm>
16  
17 using namespace std;
18 typedef pair<int, int> PAIR;
19 typedef long long ll;
20  
21 map<int, vector<PAIR> > M;
22 map<int, vector<PAIR> > ::iterator now;
23 vector <PAIR> ::iterator I;
24 multiset <int, greater<int> > S;
25 ll ans;
26 int n, L, R, H, K;
27  
28 inline int read(int &x){
29     int sgn = 1; x = 0;
30     char ch = getchar();
31     while (ch < '0' || ch > '9'){
32         if (ch == '-') sgn = -1;
33         ch = getchar();
34     }
35     while (ch >= '0' && ch <= '9'){
36         x = x * 10 + ch - '0';
37         ch = getchar();
38     }
39     x *= sgn;
40 }
41  
42 int main(){
43     read(n);
44     for (int i = 1; i <= n; ++i){
45         read(L), read(R), read(H);
46         M[L].push_back(make_pair(H, 1));
47         M[R].push_back(make_pair(H, 0));
48     }
49     L = 0;
50     for (now = M.begin(); now != M.end(); ++now){
51         R = now -> first;
52         if (!S.empty()) ans += (ll) (R - L) * (*S.begin());
53         L = R;
54         for (I = (now -> second).begin(); I != (now -> second).end(); ++I){
55             K = I -> first;
56             if (I -> second) S.insert(K);
57             else S.erase(S.find(K));
58         }
59     }
60     printf("%lld\n", ans);
61     return 0;
62 }
View Code

 (反正蒟蒻自己是不会写的。。。真是太神了!!!)

posted on 2014-10-21 15:26  Xs酱~  阅读(735)  评论(0编辑  收藏  举报