Codeforces 777E（贪心+单调栈）

Codeforces 777E Hanoi Factory

题目:
Hanoi Factory
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Of course you have heard the famous task about Hanoi Towers, but did you know that there is a special factory producing the rings for this wonderful game? Once upon a time, the ruler of the ancient Egypt ordered the workers of Hanoi Factory to create as high tower as possible. They were not ready to serve such a strange order so they had to create this new tower using already produced rings.

There are n rings in factory’s stock. The i-th ring has inner radius ai, outer radius bi and height hi. The goal is to select some subset of rings and arrange them such that the following conditions are satisfied:

Outer radiuses form a non-increasing sequence, i.e. one can put the j-th ring on the i-th ring only if bj ≤ bi.
Rings should not fall one into the the other. That means one can place ring j on the ring i only if bj > ai.
The total height of all rings used should be maximum possible.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of rings in factory’s stock.

The i-th of the next n lines contains three integers ai, bi and hi (1 ≤ ai, bi, hi ≤ 109, bi > ai) — inner radius, outer radius and the height of the i-th ring respectively.

Output
Print one integer — the maximum height of the tower that can be obtained.

Examples
input
3
1 5 1
2 6 2
3 7 3
output
6
input
4
1 2 1
1 3 3
4 6 2
5 7 1
output
4
Note
In the first sample, the optimal solution is to take all the rings and put them on each other in order 3, 2, 1.

In the second sample, one can put the ring 3 on the ring 4 and get the tower of height 3, or put the ring 1 on the ring 2 and get the tower of height 4.

题意：
有n个圆环，分别给出每个圆环的内径，外径，厚度，要求按照外径从大到小往上摞，若（相邻的圆环，上面的外径必须大于下层元环的内径）问最多可以摞多高。

首先对所有圆环按照外径从大到小排序，外径相同的按照内径从大到小排序，排完序后，由于圆环内径一旦大于下一个圆环的外径，则这个圆环之后的圆环都不能放在其上，所以可以用一个单调栈，用maxn记录当前栈内高度，一旦发现下一个圆环外径比栈顶圆环内径小，泽吧栈内元素pop知道能放下下一个圆环，然后用ans比对记录最大的一个。

AC代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <string>
#include <set>
#include <queue>
#include <stack>
#include <vector>
using namespace std;
struct xx
{
    __int64 a,b,c;
}num[100005];
stack<int> s;
int cmp(xx m,xx n)
{
    if(m.b==n.b) return m.a>n.a;
    else return m.b>n.b;
}
bool check (int i)
{
    if(s.empty()) return true;
    int x=s.top();
    if(num[x].a>=num[i].b) return false;
    else return true;
}
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF){
        for(int i=0;i<n;i++){
            cin>>num[i].a>>num[i].b>>num[i].c;
        }
        sort(num,num+n,cmp);
        __int64 ans=0;
        __int64 now=0;
        while(!s.empty())
            s.pop();
        for(int i=0;i<n;i++){
            while(!check(i)){
                int x=s.top();
                s.pop();
                now-=num[x].c;

            }
            s.push(i);
            now+=num[i].c;
            ans=max(now,ans);
        }
        printf("%I64d\n",ans);
    }
    return 0;
}