hdoj--1384--Intervals（差分约束）

Intervals

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 3559    Accepted Submission(s): 1303

Problem 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 endpoints 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 <= 50 000) - 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 <= 50 000 and 1 <= ci <= bi - ai + 1.

Process to the end of file.

 

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

 

Author

1384

 

Recommend

Eddy   |   We have carefully selected several similar problems for you:  1529 1531 1548 1534 1317 

题意：给出n个区间的左右端点，和这个区间内至少存在的在集合s中的点的个数，让你求集合s中最少有多少个点

题解：重在找到差分约束的约束条件，将约束条件转化为xj-xi<=k的形式，然后建立一条从i到j权值为k的边；

设maxl为区间的左端点，maxr为区间的右端点，S[i] 表示集合Z里面的元素在区间[0, i ]的个数，Maxl,Maxr分别表示所有区间里面的最左端和最右端，dist[]数组存储源点到某点的最短路。则由题意得限制条件

一 S[right] -  S[left-1] >= least 即[left, right]区间个数不小于least，转换得S[left-1] - S[right] <= least；

二 0 <= S[i] - S[i-1] <= 1转换得 S[i-1] - S[i] <= 0 && S[i] - S[i-1] <= 1。

第二个条件题中并没有给出，需要自己推导，因为仅仅靠题中的条件无法构建一个连通图，也就无法求最短路，因为s[i]表示的是集合Z里面的元素在区间[0, i ]的个数所以s[i]至多比s[i-1]大一也可能相等

然后根据限制条件建图

转化问题：题目需要求的是S[Maxr] - S[Maxl-1] >= ans 即S[Maxl-1] - S[Maxr] <= -ans。 若以Maxr为源点 ，而-ans就为Maxr到Maxl-1的最短路径的相反数，即-dist[Maxl-1]。

#include<stdio.h>
#include<string.h>
#include<queue>
#include<iostream>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f
int head[100100],vis[1000100],n,cnt;
int dis[100100];
int l,r;
struct node
{
	int u,v,val;
	int next;
}edge[1001000];
void init()
{
	cnt=0;
	memset(head,-1,sizeof(head));
	l=INF;
	r=-1;
}
void add(int u,int v,int val)
{
	node E={u,v,val,head[u]};
	edge[cnt]=E;
	head[u]=cnt++;
}
void getmap()
{
	int a,b,c;
	for(int i=0;i<n;i++)
	{
		cin>>a>>b>>c;
		l=min(a,l);
		r=max(r,b);
		add(b,a-1,-c);
	}
	for(int i=l;i<=r;i++)
	{
		add(i,i-1,0);
		add(i-1,i,1);
	}
}
void SPFA()
{
	for(int i=l-1;i<=r;i++)
	dis[i]=i==r?0:INF; 
	memset(vis,0,sizeof(vis));
	queue<int>q;
	q.push(r);
	vis[r]=1;
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		vis[u]=0;
		for(int i=head[u];i!=-1;i=edge[i].next)
		{
			node E=edge[i];
			if(dis[E.v]>dis[E.u]+edge[i].val)
			{
				dis[E.v]=dis[E.u]+edge[i].val;
				if(!vis[E.v])
				{
					vis[E.v]=1;
					q.push(E.v);
				}
			}
		}
	}
	cout<<-dis[l-1]<<endl;
}
int main()
{
	while(scanf("%d",&n)!=EOF)
	{
		init();
		getmap();
		SPFA();
	}
	return 0;
}