HDOJ 题目1384 Intervals(差分约束)

Intervals

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

Total Submission(s): 3029    Accepted Submission(s): 1117


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

 

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 题目大意：

在每个区间[ai,bi]上至少选择ci个元素，构成一个集合S，使集合S中的元素最少；
分析：f(a) 表示在区间 [0,a] 上选择了一些元素；
则：f(b)-f(a-1)>=ci;
   0<=f(a)-f(a-1)<=1;
ac代码

#include<stdio.h>
#include<string.h>
#include<string>
#include<queue>
#include<iostream>
#define INF 0xfffffff
using namespace std;
int head[500500],vis[500500],dis[500500],id,l,r;
struct s
{
int to,w,next;
}node[500050];
void add(int u,int v,int w)
{
node[id].w=w;
node[id].to=v;
node[id].next=head[u];
head[u]=id++;
}
void spfa()
{
int i,u,v;
queue<int>q;
for(i=l;i<=r;i++)
{
dis[i]=INF;
vis[i]=0;
}
q.push(r);
dis[r]=0;
vis[r]=1;
while(!q.empty())
{
int u=q.front();
q.pop();
vis[u]=0;
for(i=head[u];i!=-1;i=node[i].next)
{
int v=node[i].to;
int temp=dis[u]+node[i].w;
if(temp<dis[v])
{
dis[v]=temp;
if(!vis[v])
{
q.push(v);
vis[v]=1;
}
}
}
}
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
int i;
l=INF;
r=0;
memset(head,-1,sizeof(head));
id=0;
for(i=0;i<n;i++)
{
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
add(b+1,a,-c);
if(a<l)
l=a;
if(b>r)
r=b;
}
r++;
for(i=l;i<r;i++)
{
add(i+1,i,0);
add(i,i+1,1);
}
spfa();
printf("%d\n",-dis[l]);
}
}