Intervals poj1201(差分约束)

最新推荐文章于 2019-10-01 10:53:10 发布

SDUTyangkun

最新推荐文章于 2019-10-01 10:53:10 发布

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本文介绍了一个关于区间覆盖的问题，并提供了一种通过编程解决该问题的方法。具体来说，需要找到一个整数集合，使得该集合与每个给定的闭合区间至少有指定数量的共同元素，目标是最小化这个集合的大小。

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Intervals

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 25756		Accepted: 9854

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

Source

Southwestern Europe 2002

注意！！！！原点是dis[max] 而且由于本例题s0-s11<=-M,既要求原点s11到s0的最短路径长度为-M

假设最短路径求得各顶点到s11原点距离保存到dis[]里面那么-M=dis[0]-dis[11],M=dis[11]-dis[0]

#include<stdio.h>
#include<string.h>
#include<queue>
#define maxn 55000
#define INF 0x3f3f3f3f
#define max(a,b) a>b?a:b
#define min(a,b) a>b?b:a
using namespace std;

struct node
{
    int v,w,next;
}e[maxn*3];
int dis[maxn],vis[maxn],head[maxn];
int cnt;
void init()
{
    cnt=0;
    memset(head,-1,sizeof(head));
    memset(vis,0,sizeof(vis));
}
void add(int u,int v,int w)
{
    e[cnt].v=v;
    e[cnt].w=w;
    e[cnt].next=head[u];
    head[u]=cnt++;
}
void spfa(int mi,int ma)
{
    int i;
    for(i=mi;i<=ma;i++)
    {
         dis[i]=INF;
    }

    dis[ma]=0;
    vis[ma]=1;
    queue<int>q;
    q.push(ma);

    while(!q.empty())
    {

        int u=q.front();q.pop();
        vis[u]=0;
        for(int i=head[u];i!=-1;i=e[i].next)
        {
            int v=e[i].v;
            int w=e[i].w;
            if(dis[v]>dis[u]+w)
            {
                dis[v]=dis[u]+w;
                if(!vis[v])
                {
                    q.push(v);
                    vis[v]=1;
                }
            }
        }
    }
}
int main()
{
    int n;
    int i,j;
    while(~scanf("%d",&n))
    {
        int a,b,x;
        int MIN=INF;
        int MAX=-1;
        init();
        for(i=1;i<=n;i++)
        {
            scanf("%d%d%d",&a,&b,&x);
            MIN=min(a,MIN);
            MAX=max(b+1,MAX);
            add(b+1,a,-x);
        }

        for(i=0;i<MAX;i++)
        {
            add(i+1,i,0);
            add(i,i+1,1);
        }

        spfa(MIN,MAX);
        printf("%d\n",dis[MAX]-dis[MIN]);
    }
}