POJ 1201 Intervals

Intervals

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 19359		Accepted: 7308

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

5
3 7 3
8 10 3
6 8 1
1 3 1
10 11 1

Sample Output

Source

Southwestern Europe 2002

链式前向星SPFA+差分约束

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>

using namespace std;

const int MAXN=550000;
const int INF=0x3f3f3f3f;

typedef struct
{
    int next,to,w;
}Edge;

int maxv=-1;
Edge E[MAXN];
int Size,Adj[MAXN];

void Init()
{
    Size=0;
    memset(Adj,-1,sizeof(Adj));
}

void Add_Edge(int u,int v,int c)
{
    E[Size].to=v;
    E[Size].w=c;
    E[Size].next=Adj;
    Adj=Size++;
}
int dist[MAXN]; bool inQue[MAXN];
void SPFA()
{
    queue<int> q;
    for(int i=0;i<=maxv;i++)
    {
        dist=-INF;
        inQue=false;
    }
    dist[0]=0; inQue[0]=true;
    q.push(0);
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        for(int i=Adj;~i;i=E.next)
        {
            if(dist+E.w>dist[E.to])
            {
                dist[E.to]=dist+E.w;
                if(!inQue[E.to])
                {
                    inQue[E.to]=true;
                    q.push(E.to);
                }
            }
        }
        inQue=false;
    }
    printf("%d ",dist[maxv]);
}

int main()
{
    int T;
    scanf("%d",&T);
    Init();
    while(T--)
    {
        int a,b,c;
        scanf("%d%d%d",&a,&b,&c);
        if(b+1>maxv) maxv=b+1;
        Add_Edge(a,b+1,c);
    }
    for(int i=0;i<=maxv;i++)
    {
        Add_Edge(i+1,i,-1);
        Add_Edge(i,i+1,0);
    }
    SPFA();
    return 0;
}

* This source code was highlighted by YcdoiT. ( style: Codeblocks )