poj 1201 Intervals（差分约束，最长路）

本文链接：https://blog.youkuaiyun.com/acm_cxq/article/details/52001764

这篇博客主要解析了POJ 1201题目，涉及差分约束系统和最长路径问题。通过设置s[i]表示0到i区间Z集合的数量，并利用s[b[i]] - s[a[i-1]] >= c[i]的条件，结合每个位置上的Z数变化不超过1，采用SPFA算法求解最长路。起点可以是超级源点或者所有区间中最左点。

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Intervals

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 24955		Accepted: 9497

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.

Sample Input

Sample Output

题意：给n个区间，每个区间上至少要用数字集Z的c[i]，问Z最小为多少个

思路：设s[i]表示区间0~i上Z集合中的个数于是有s[b[i]]-s[a[i-1]]>=c[i]

还有注意每一个位置上要么有Z集合中的数，要么没有，所以还有0<=s[i]-s[i-1]<=1

最后spfa跑一遍最长路即可

注意起点可以是超级源点，也可以是所有区间里面最左边的一个点

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;
#define N 50050
#define INF 999999999
struct Edge
{
    int v,next,w;
}edge[N*20];
int cnt,head[N];
int d[N],vis[N];
void init()
{
    cnt=0;
    memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w)
{
    edge[cnt].v=v;edge[cnt].w=w;
    edge[cnt].next=head[u];
    head[u]=cnt++;
}
int spfa(int s,int t)
{
    memset(vis,0,sizeof(vis));
    for(int i=s;i<=t;i++)
        d[i]=i==s?0:-INF;
    vis[s]=1;
    queue<int>que;
    que.push(s);
    while(!que.empty())
    {
        int u=que.front();
        que.pop();
        vis[u]=0;
        for(int i=head[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].v,w=edge[i].w;
            if(d[v]<d[u]+w)
            {
                d[v]=d[u]+w;
                if(!vis[v])
                {
                    vis[v]=1;
                    que.push(v);
                }
            }
        }
    }
    return d[t];
}
int main()
{
    int n,p,q;
    int s,t,w;
    while(~scanf("%d",&n))
    {
        init();
        int maxn=-1,minn=INF;
        while(n--)
        {
            scanf("%d %d %d",&s,&t,&w);
            addedge(s-1,t,w);
            maxn=max(maxn,t);
            minn=min(minn,s-1);
        }
        for(int i=minn+1;i<=maxn;i++)
        {
            addedge(i,i-1,-1);
            addedge(i-1,i,0);
        }
        printf("%d\n",spfa(minn,maxn));
    }
    return 0;
}