poj1201 Intervals

最新推荐文章于 2024-08-17 03:35:41 发布

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最新推荐文章于 2024-08-17 03:35:41 发布

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分类专栏：图论 poj 文章标签：差分约束系统

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本文介绍了一种使用差分约束系统解决区间覆盖问题的方法。通过将原始问题转化为更易求解的形式，即利用前缀和优化，使得问题可以通过构建特定的图结构来解决。文章详细展示了如何设置限制条件，并给出了完整的C++实现代码。

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.

差分约束系统。
对于a[i..j]>=c，利用前缀和优化成s[j]-s[i-1]>=c，就可以用差分约束系统了。
注意本题的特殊性，还应该加上限制条件s[i]>=s[i-1]，s[i]<=s[i-1]+1。

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int dis[50010],fir[50010],ne[200010],to[200010],len[200010];
queue<int> q;
void add(int num,int f,int t,int l)
{
    ne[num]=fir[f];
    fir[f]=num;
    to[num]=t;
    len[num]=l;
}
int main()
{
    int i,j,k,m,n,p,x,y,z,ans=0,u,v;
    scanf("%d",&n);
    for (i=1;i<=n;i++)
    {
        scanf("%d%d%d",&x,&y,&z);
        add(i,y,x-1,-z);
    }
    for (i=1;i<=50000;i++)
    {
        add(i*2+n,i,i-1,0);
        add(i*2+1+n,i-1,i,1);
    }
    for (i=0;i<=50000;i++)
      q.push(i);
    while (!q.empty())
    {
        u=q.front();
        q.pop();
        for (i=fir[u];i;i=ne[i])
        {
            v=to[i];
            if (dis[u]+len[i]<dis[v])
            {
                dis[v]=dis[u]+len[i];
                q.push(v);
            }
        }
    }
    for (i=0;i<=50000;i++)
      ans=max(ans,-dis[i]);
    printf("%d\n",ans);
}