[poj 1201]Intervals 差分约束

发布时间：2020-12-13 21:07:07 所属栏目：PHP教程来源：网络整理

导读：题目：http://poj.org/problem?id=1201 Intervals Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 24502 Accepted: 9317 Description You are given n closed,integer intervals [ai,bi] and n integers c1,…,cn. Write a program that: rea

题目：http://poj.org/problem?id=1201

Intervals
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 24502 Accepted: 9317

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,n.

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

Sample Output
6

题意：要求满足bi-ai选数>=ci 中最小的集合；

思路：
如题：s[n]代表前n当选最小的集合，可得：

    s[bi]-s[ai-1]>=ci;（题）
    s[I+1]-s[I]>=0;（本身）
    s[I+1]-s[I]<=1;-->s[l]-s[I+1]>=-1（本身）

用最小值建边spfa求maxn-minn的最长路便可；

代码：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
int head[60005],val[160005];
int tot,to[160005],nex[160005];
int n;
int aa,bb,cc;
queue<int> q;
int d[60005];
int vis[60005];
void spfa(int x)
{
    memset(d,-1,sizeof(d));
    d[x]=0;
    vis[x]=1;
    q.push(x);

    while(!q.empty())
    {
        int now=q.front();
        q.pop();
        vis[now]=0;

        for(int i=head[now];i!=-1;i=nex[i])
        {
            int v=to[i];
            if(d[v]==-1||d[v]<d[now]+val[i])
            {

                d[v]=d[now]+val[i];

                if(!vis[v])
                {
                    vis[v]=1;
                    q.push(v);
                } 
            }
        }
    }
}
void add(int x,int y,int v)
{
    int tmp=head[x];
    head[x]=++tot;
    nex[tot]=tmp;
    to[tot]=y;
    val[tot]=v; 
} 

int main()
{
    scanf("%d",&n);
    int minn=10000000;
    int maxn=-10000000;
    memset(head,sizeof(head));
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d%d",&aa,&bb,&cc);
        aa--;
        add(bb,aa,cc);
        minn=min(aa,minn);
        maxn=max(maxn,bb);
    }
    for(int i=minn;i<maxn;i++)
    {
        add(i+1,i,0);
        add(i,i+1,-1);
    }
    spfa(maxn);
    printf("%dn",d[minn]);

}

（编辑：李大同）

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