poj 1716 Integer Intervals(差分约束||贪心)(中等)

Integer Intervals

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 13229		Accepted: 5613

Description

An integer interval [a,b], a < b, is a set of all consecutive integers beginning with a and ending with b.

Write a program that: finds the minimal number of elements in a set containing at least two different integers from each interval.
Input

The first line of the input contains the number of intervals n, 1 <= n <= 10000. Each of the following n lines contains two integers a, b separated by a single space, 0 <= a < b <= 10000. They are the beginning and the end of an
interval.

Output
Output the minimal number of elements in a set containing at least two different integers from each interval.

Sample Input

4
3 6
2 4
0 2
4 7

Sample Output

4

思路：

这题贪心和差分约束都可以，在poj上交wa了6变，不是我的问题，是系统傻逼。。。这类型的题要写出自己的风格

代码：

#include<iostream>
using namespace std;
//#define maxn 10001
#define inf 20000

struct edge
{
int u,v;
}edge[10001];
int dist[10001];
int l,r;
int n;

int main(int i)
{
while(cin>>n)
{
l=inf,r=0;
for(i=0;i<n;i++)
{
int a,b;
cin>>a>>b;
edge[i].u=a,edge[i].v=b+1;
if(edge[i].u<l) l=edge[i].u;
if(edge[i].v>r) r=edge[i].v;
dist[i]=0;
}
int flag=1;
while(flag)
{
flag=0;
for(i=0;i<n;i++)
{
if(dist[edge[i].u]>dist[edge[i].v]-2)
{
dist[edge[i].u]=dist[edge[i].v]-2;
flag=1;
}
}
for(i=l;i<l;i++)
{
if(dist[i+1]>dist[i]+1)
{
dist[i+1]=dist[i]+1;
flag=1;
}
}
for(i=r-1;i>=l;i--)
{
if(dist[i]>dist[i+1])
{
dist[i]=dist[i+1];
flag=1;
}
}
}
cout<<dist[r]-dist[l]<<endl;
}
return 0;
}

贪心：

这个模型建的好，速度更快

//260K	141MS
#include<iostream>
#include<algorithm>
using namespace std;

typedef struct edge
{
int s,e;
}edge;
int n;
int e1,e2;

bool cmp(edge a,edge b)
{
if(a.e!=b.e)
return a.e<b.e;
return a.s<b.s;
}

int main()
{
while(cin>>n)
{
int sum=2;
edge *t=new edge
;
for(int i=0;i<n;i++)
cin>>t[i].s>>t[i].e;
sort(t,t+n,cmp);
e1=t[0].e-1,e2=t[0].e;
for(int k=1;k<n;k++)
{
if(e1>=t[k].s)
continue;
else if(e1<t[k].s && e2>=t[k].s)
{
e1=e2;
e2=t[k].e;
sum++;
}
else
{
sum+=2;
e1=t[k].e-1;
e2=t[k].e;
}
}
cout<<sum<<endl;
delete t;
}
return 0;
}