【POJ 1201 Intervals】

Time Limit: 2000MS
Meamory Limit: 65536K

Total Submissions: 27949
Accepted: 10764

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

【题解】

      ①这是一个差分约束模型。

      ②本题约束条件基于前缀和sum数组。

      ③外加相邻点的前缀和关系（最多相差1）。

#include<queue>
#include<stdio.h>
#include<algorithm>
#define inf 1000000007
#define go(i,a,b) for(int i=a;i<=b;i++)
#define fo(i,a,x) for(int i=a[x],v=e[i].v;i;i=e[i].next,v=e[i].v)
const int N=50010;
using namespace std;
bool inq[N];queue<int>q;
struct E{int v,next,w;}e[N<<1];
int n,a[N],b[N],c[N],head[N],k=1,S=1e9,T,d[N];
void ADD(int u,int v,int w){e[k]=(E){v,head[u],w};head[u]=k++;}

void Build()
{
	go(i,1,n)scanf("%d %d %d",a+i,b+i,c+i);
	go(i,1,n)S=min(S,a[i]-1),T=max(T,b[i]);
	go(i,1,n)ADD(a[i]-1,b[i],c[i]);
	go(i,S+1,T)ADD(i,i-1,-1);
	go(i,S,T-1)ADD(i,i+1,0);
}

void SPFA()
{
	go(i,S,T)d[i]=-inf;d[S]=0;q.push(S);int u;
	while(!q.empty()){inq[u=q.front()]=0;q.pop();fo(i,head,u)
	if(d[u]+e[i].w>d[v])d[v]=d[u]+e[i].w,!inq[v]?q.push(v),inq[v]=1:1;}
}

int main()
{
	scanf("%d",&n);Build();
	SPFA();printf("%d\n",d[T]);
	return 0;
}//Paul_Guderian

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