Intervals

Time Limit: 2000MS		Memory Limit: 65536K

Description

You are given n closed, integer intervals [ai, bi] and n integers c1, ..., cn. 

Write a program that: 

reads the number of intervals, their end points and integers c1, ..., cn from the standard input, 

computes the minimal size of a set Z of integers which has at least ci common elements with interval [ai, bi], for each i=1,2,...,n, 

writes the answer to the standard output. 

Input

The first line of the input contains an integer n (1 <= n <= 50000) -- the number of intervals. The following n lines describe the intervals. The (i+1)-th line of the input contains three integers ai, bi and ci separated by single spaces and such that 0 <= ai <= bi <= 50000 and 1 <= ci <= bi - ai+1.

Output

The output contains exactly one integer equal to the minimal size of set Z sharing at least ci elements with interval [ai, bi], for each i=1,2,...,n.

Sample Input

53 7 38 10 36 8 11 3 110 11 1

Sample Output

6

Source

 

题目大意：给出n个区间[a,b]，每个区间指定要选出c个数，求满足所有的区间最少要选出多少个数

poj 1716的弱化版

在此不再详细解读

可参考 

#include
      
       #include
       
        #include
        
         #include
         
          #define N 50001using namespace std;int n,minn=50002,maxn;queue
          
           q;struct node{    int to,next,w;}e[N*3];int dis[N],front[N],tot;bool v[N];void add(int u,int v,int w){    e[++tot].to=v;e[tot].next=front[u];front[u]=tot;e[tot].w=w;}int main(){    scanf("%d",&n);    int x,y,z;    for(int i=1;i<=n;i++)    {        scanf("%d%d%d",&x,&y,&z);y++;        add(x,y,z);        minn=min(x,minn);maxn=max(maxn,y);    }    for(int i=minn;i