poj 1716 Integer Intervals

Posted 2020-09-03 日拱一卒 功不唐捐

 Integer Intervals

http://poj.org/problem?id=1716

Time Limit: 1000MS		Memory Limit: 10000K

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

Source

CEOI 1997

题目大意：在指定的几个区间中选出最少的数，使每个区间至少包含这些数中的2个，求最少多少数

解法一：贪心

把所有的区间按右端点从小到大排序

初始时，假设这两个数x，y为第一个区间最右边的2个数，x<=y，ans=2

如果下一个区间左端点<=x，跳过

如果下一个区间的左端点在>x 且<=y，令x=y，y=当前区间右端点，ans++

如果下一个区间的左端点>y   x，y为当前区间最右边的两个数，ans+2

#include<cstdio>
#include<algorithm>
using namespace std;
int n,ans;
struct node
{
    int a,b;
}e[10001];
bool cmp(node p,node q)
{
    return p.b<q.b;
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d%d",&e[i].a,&e[i].b);
    sort(e+1,e+n+1,cmp);
    int x=e[1].b-1,y=x+1;
    ans=2;
    for(int i=2;i<=n;i++)
    {
        if(e[i].a<=x) continue;
        if(e[i].a<=y) {x=y;y=e[i].b;ans++;continue;}
        x=e[i].b-1;y=x+1;ans+=2;
    }
    printf("%d",ans);
}

View Code

解法二：差分约束

某大佬差分约束详解：http://www.cppblog.com/menjitianya/archive/2015/11/19/212292.html

若d[x]-d[y]<=/>= c ,则由y向x连一条权值为c的边

设d[i]为从1号点到i最少选出的数的个数

根据题目要求，对于区间[l,r],得约束条件1:d[r]-d[l-1]>=2

为保证连续性，隐含约束条件2：0<=d[i]-d[i-1]<=1

即每个点要么不选，要么只能选一次

设总区间[s,t]

那么目的就是 d[t]-d[s]>=ans

即d[t]>=ans+d[s]，所以spfa跑最长路

因为题目中，输入数据端点可能有0，所以我们将所有的数整体后移一位

#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
#define N 10002
using namespace std;
int n,minn=10002,maxn;
queue<int>q;
struct node
{
    int to,next,w;
}e[N*3];
int dis[N],front[N],tot;
bool v[N];
void add(int u,int v,int w)
{
    e[++tot].to=v;e[tot].next=front[u];front[u]=tot;e[tot].w=w;
}
int main()
{
        int x,y;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&x,&y);y++;
            add(x,y,2);
            minn=min(x,minn);maxn=max(maxn,y);
        }
        for(int i=minn;i<maxn;i++) 
        {
            add(i,i+1,0);add(i+1,i,-1);
        }
        memset(dis,-1,sizeof dis);
        q.push(minn);v[minn]=true;dis[minn]=0;
        while(!q.empty())
        {
            int now=q.front();q.pop();v[now]=false;
            for(int i=front[now];i;i=e[i].next)
            {
                int to=e[i].to;
                if(dis[to]<dis[now]+e[i].w)
                {
                    dis[to]=dis[now]+e[i].w;
                    if(!v[to])
                    {
                        v[to]=true;
                        q.push(to);
                    }
                }
            }
        }
        printf("%d",dis[maxn]);
}

View Code

一个错误：N=10001，因为全体后移了一位，所以N最少是10002