(差分约束系统) poj 1201

Intervals

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 22781		Accepted: 8613

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：有n个区间，每个区间有3个值，ai，bi，ci代表，在区间[ai,bi]上至少要选择ci个整数点，ci可以在区间内任意取不重复的点
现在要满足所有区间的自身条件，问最少选多少个点

解题思路：
差分约束的思想：可以肯定的是s[bi]-s[ai-1]>=ci; 为什么要ai-1，是因为ai也要选进来
在一个是s[i]-s[i-1]<=1;
s[i]-s[i-1]>=0
所以整理上面三个式子可以得到约束条件：
①s[ai-1]-s[bi] <= -ci
②s[i]-s[i-1] <= 1
③s[i-1]-s[i] <= 0

 

#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<climits>
using namespace std;
int n,minn,maxx,cnt;
bool vis[50010];
int dist[50010];
struct node
{
    int to,len,next;
}e[210000];
int head[50010];
void add(int u,int v,int len)
{
    e[++cnt].to=v;
    e[cnt].len=len;
    e[cnt].next=head[u];
    head[u]=cnt;
}
void spfa()
{
    queue<int> q;
    for(int i=minn;i<=maxx;i++)
    {
        vis[i]=0;
        dist[i]=INT_MAX;
    }
    dist[maxx]=0;
    vis[maxx]=1;
    q.push(maxx);
    while(!q.empty())
    {
        int x=q.front();
        q.pop();
        vis[x]=0;
        for(int i=head[x];i;i=e[i].next)
        {
            int v=e[i].to;
            int w=e[i].len;
            if(dist[v]>dist[x]+w)
            {
                dist[v]=dist[x]+w;
                if(!vis[v])
                {
                    q.push(v);
                    vis[v]=1;
                }
            }
        }
    }
}
int main()
{
    scanf("%d",&n);
    cnt=1;
    minn=INT_MAX,maxx=-INT_MAX;
    for(int i=1;i<=n;i++)
    {
        int a,b,c;
        scanf("%d%d%d",&a,&b,&c);
        minn=min(a,minn);
        maxx=max(maxx,b+1);
        add(b+1,a,-c);
    }
    for(int i=minn;i<maxx;i++)
    {
        add(i+1,i,0);
        add(i,i+1,1);
    }
    spfa();
    printf("%d\n",-dist[minn]);
    return 0;
}