poj 1201 Intervals

Intervals

http://poj.org/problem?id=1201

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

5
3 7 3
8 10 3
6 8 1
1 3 1
10 11 1

Sample Output

6

Source

Southwestern Europe 2002

 

题目大意：给出n个区间[a,b]，每个区间至少要选出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 http://www.cnblogs.com/TheRoadToTheGold/p/6529744.html

#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
#define N 50001
using namespace std;
int n,minn=50002,maxn;
queue<int>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<maxn;i++) 
    {
        add(i,i+1,0);add(i+1,i,-1);
    }
    memset(dis,-1,sizeof(dis));
    q.push(minn);v[minn]=true;dis[minn]=0;
    while(!q.empty())
    {
        int now=q.front();q.pop();v[now]=false;
        for(int i=front[now];i;i=e[i].next)
        {
            int to=e[i].to;
            if(dis[to]<dis[now]+e[i].w)
            {
                dis[to]=dis[now]+e[i].w;
                if(!v[to])
                {
                    v[to]=true;
                    q.push(to);
                }
            }
        }
    }
    printf("%d",dis[maxn]);
}