bzoj 1664 （贪心）

 [Usaco2006 Open]County Fair Events 参加节日庆祝

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 487  Solved: 344
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Description

Farmer John has returned to the County Fair so he can attend the special events (concerts, rodeos, cooking shows, etc.). He wants to attend as many of the N (1 <= N <= 10,000) special events as he possibly can. He's rented a bicycle so he can speed from one event to the next in absolutely no time at all (0 time units to go from one event to the next!). Given a list of the events that FJ might wish to attend, with their start times (1 <= T <= 100,000) and their durations (1 <= L <= 100,000), determine the maximum number of events that FJ can attend. FJ never leaves an event early.

 

有N个节日每个节日有个开始时间,及持续时间. 牛想尽可能多的参加节日,问最多可以参加多少. 注意牛的转移速度是极快的,不花时间.

Input

* Line 1: A single integer, N.

* Lines 2..N+1: Each line contains two space-separated integers, T and L, that describe an event that FJ might attend.

Output

* Line 1: A single integer that is the maximum number of events FJ can attend.

Sample Input

7
1 6
8 6
14 5
19 2
1 8
18 3
10 6

INPUT DETAILS:

Graphic picture of the schedule:
11111111112
12345678901234567890---------这个是时间轴.
--------------------
111111 2222223333344
55555555 777777 666

这个图中1代表第一个节日从1开始,持续6个时间,直到6.

Sample Output

4

 

这里很容易贪心证明，不需要去O（n^2）去跑，很容易被卡。

就是这里的，想象成许多线段，如果相交，若当前线段右端比以前线段右端小，那么就赋值为现在那条线段，因为这样只会更优。

如果不想交，直接ans+1

 

划水

#include<cstring>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<cstdio>

#define N 10007
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
    while(ch<='9'&&ch>='0')
    {
        x=(x<<3)+(x<<1)+ch-'0';
        ch=getchar();
    }
    return x*f;
}

int n,ans;
struct Node
{
    int x,y;
}a[N];

bool cmp(Node x,Node y)
{
    return x.x<y.x;
}
int main()
{
    n=read();
    for (int i=1;i<=n;a[i].x=read(),a[i].y=a[i].x+read()-1,i++);
    sort(a+1,a+n+1,cmp);
    
    for (int i=1,r=0;i<=n;i++)
    {
        if (a[i].x>r) ans++,r=a[i].y;
        else if (a[i].y<r) r=a[i].y;
    }
    printf("%d",ans);
}