HDU2652Warching TV （二分 dp）

Warching TV
Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 223    Accepted Submission(s): 80
Problem Description
Lemon likes warching TV very much. When winter holiday comes, it is a good time, isn’t it. There are lots of TV program listed on the paper. Every TV program has its start time ,end time, and the happiness value that will add Lemon’s happiness and it depends on Lemon’s taste.
 
Input
There are many test cases. Please process to end of file. Each test case starts with one integer N (1 <= N <= 100000) which indicates the size of the list of the TV program. Then N lines follow, each line contains three integers s, e and v. s, e indicate that the TV program is during [s, e](1 <= s <= e <= 1000000). v indicates that after warching the TV program will add Lemon v happyiness(1 <= v <= 1000). Once Lemon choose a TV program, he must finish warching the TV program.
 
Output
Print the maximum happiness value that Lemon will get.
 
Sample Input

2
1 2 1
3 4 2
2
1 2 1
2 4 2

 
Sample Output

3
2

 
Author
lemon
 
Source
决战龙虎门
 
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 题意：给了n个节目的开始时间，结束时间，和看完节目可收获的开心值，一个节目结束的时间必须大于下一个节目开始的时间，计算可以可以收获的最大开心值和。

思路:dp，二分

注意：把结构体定义在外面，不然会超时

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int N = 100010;
struct node
{
    int s,e,v;
} nodes[N];    //定义到外面，可以节省运行时间，不然会超时，，
bool comp(node a,node b)
{
    if(a.e!=b.e)
    return a.e<b.e;
    return a.s<b.s;
}
int value[N];
int dp[N];
int main()
{
    int n;
  
    while(~scanf("%d",&n))
    {
        memset(dp,0,sizeof(dp));
        for(int i=0;i<n;i++)
    {
        scanf("%d%d%d",&nodes[i].s,&nodes[i].e,&nodes[i].v);

    }
    sort(nodes,nodes+n,comp);
    int left,mid,right;
    dp[0]=nodes[0].v;
    for(int i=1;i<n;i++)
    {
        dp[i]=nodes[i].v;
        left=0;
        right=i-1;
        int ans=-1;
        mid=(left+right)/2;
        while(left<=right)   //二分找到最省时的节目播放方式
        {
            mid=(left+right)/2;
        if(nodes[mid].e<nodes[i].s)
        {
            ans=mid;
            left=mid+1;
        }
        else
        right=mid-1;
        }
        if (ans != -1)   //如果ans的值为-1，则说明第i个节目结束后没有节目可看了。
   {
    dp[i] = max(dp[i], dp[ans] + nodes[i].v);
   }
   dp[i] = max(dp[i], dp[i - 1]);
        //dp[i]=max(dp[i-1],max(dp[i],dp[mid]+nodes[i].v));   这样写没有考虑到看完当前节目后没有可开始的节目的情况，输出结果就会是开心值的简单累加
    }
    printf("%d\n",dp[n-1]);
    }
    return 0;
}