[poj 1201]Intervals 差分约束

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, 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
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中选最小的集合，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的最长路即可；

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,-1,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("%d\n",d[minn]);}