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POJ 1201 Intervals

2016-08-02 22:09 316 查看
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.

【题目分析】

同poj1716一模一样,直接复制AC

【代码】

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
#define inf 0x3f3f3f3f
using namespace std;
int h[400001],fr[400001],ne[400001],to[400001],w[400001],en=0,d[400001];
int inq[400001];
int n,m,maxx;
inline void add(int a,int b,int c)
{to[en]=b;ne[en]=h[a];fr[en]=a;w[en]=c;h[a]=en++;}
inline void bfs()
{
for (int i=0;i<=maxx;++i) d[i]=-inf;
queue<int>q;
q.push(0);
inq[0]=1;d[0]=0;
while (!q.empty())
{
int x=q.front();q.pop();
inq[x]=0;
for (int i=h[x];i>=0;i=ne[i])
{
if (d[x]+w[i]>d[to[i]])
{
d[to[i]]=d[x]+w[i];
if (!inq[to[i]])
{
q.push(to[i]);
inq[to[i]]=1;
}
}
}
}
cout<<d[maxx]<<endl;
}
int main()
{
cin>>n;
maxx=-1;
memset(h,-1,sizeof h);
for (int i=1;i<=n;++i)
{
int a,b,c;
cin>>a>>b>>c;
add(a,b+1,c);
maxx=max(maxx,b+1);
}
for (int i=0;i<maxx;++i)
{
add(i,i+1,0);
add(i+1,i,-1);
}
bfs();
}
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