# POJ1201 Intervals(差分约束)

 Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 28416 Accepted: 10966

## 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

$S\left[ i\right] -S\left[ i-1\right] \geq 0$
$S\left[ i-1\right] -S\left[ i\right] \geq -1$

#include<cstdio>
#include<queue>
#include<cstring>
#define INF 1e8+10
using namespace std;
const int MAXN=1e6+10;
char buf[MAXN],*p1=buf,*p2=buf;
{
char c=getchar();int x=0,f=1;
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
return x*f;
}
struct node
{
int u,v,w,nxt;
}edge[MAXN];
int maxx=-INF,minn=INF;
int dis[MAXN],vis[MAXN];
inline void AddEdge(int x,int y,int z)
{
edge[num].u=x;
edge[num].v=y;
edge[num].w=z;
}
int SPFA()
{
queue<int>q;
memset(dis,-0xf,sizeof(dis));
dis[minn]=0;q.push(minn);
while(q.size()!=0)
{
int p=q.front();q.pop();
vis[p]=0;
{
if(dis[edge[i].v]<dis[p]+edge[i].w)
{
dis[edge[i].v]=dis[p]+edge[i].w;
if(vis[edge[i].v]==0)
vis[edge[i].v]=1,q.push(edge[i].v);
}
}
}
printf("%d",dis[maxx]);
}
int main()
{
#ifdef WIN32
freopen("a.in","r",stdin);
#else
#endif
for(int i=1;i<=N;i++)
{
maxx=max(y+1,maxx);
minn=min(x,minn);
}
for(int i=minn;i<=maxx-1;i++)
{
}