poj 1201 Intervals

描述

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.

输入

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.

输出

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.

题目大意：

有一个序列，题目用n个整数组合[ai,bi,ci]来描述它，[ai，bi，ci]表示在该序列中处于[ai，bi]这个区间的整数至少有ci个。如果存在这样的序列，请求出满足题目要求的最短的序列长度是多少。如果不存在则输出 -1。

输入：第一行包括一个整数n，表示区间个数，以下n行每行描述这些区间，第i+1行三个整数ai，bi，ci，由空格隔开，其中0<=ai<=bi<=50000 而且1<=ci<=bi-ai+1。

输出：一行，输出满足要求的序列的长度的最小值。

思路：

这是一道差分约束，约束方程为：

sum[bi]-sum[ai-1]>=ci,sum[i]-sum[i-1]>=0,sum[i-1]-sum[i]>=-1;（sum[i]表示从0到i满足要求的数)

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
#define inf 0x7f7f7f7f
struct node
{
int next,en,len;
}E[600000];
int head[50005],dis[50005],num,n;bool vis[50005];
void init()
{
num=0;
memset(head,-1,sizeof(head));
}
void add(int st,int en,int len)
{
E[num].en=en;E[num].len=len;
E[num].next=head[st];head[st]=num++;
}
void spfa(int st,int en)
{
int i;
for(i=0;i<=en;i++)
{
dis[i]=-inf;
vis[i]=false;
}
queue<int>q;
dis[st]=0;vis[st]=true;
q.push(st);
while(!q.empty())
{
int x=q.front();
q.pop();vis[x]=false;
for(i=head[x];i!=-1;i=E[i].next)
{
int v=E[i].en;
if(dis[v]<dis[x]+E[i].len)
{
dis[v]=dis[x]+E[i].len;
if(!vis[v])
{
vis[v]=true;
q.push(v);
}
}
}
}
}
int main()
{
int i,a,b,c,st,en;
while(scanf("%d",&n)!=EOF)
{
init();st=inf,en=-inf;
while(n--)
{

scanf("%d%d%d",&a,&b,&c);
a++,b++;//防止起点为负数
st=min(st,a-1);
en=max(en,b);
add(a-1,b,c);
}
for(i=st;i<=en;i++)
{
add(i-1,i,0);
add(i,i-1,-1);
}
spfa(st,en);
int ans=dis[en];
printf("%d\n",ans);
}
return 0;
}