poj1716 Integer Intervals（差分约束）

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Integer Intervals

Time Limit: 1000MS

Memory Limit: 10000K

Description

An integer interval [a,b], a < b, is a set of all consecutive integers beginning with a and ending with b.
Write a program that: finds the minimal number of elements in a set
containing at least two different integers from each interval.

Input

The
first line of the input contains the number of intervals n, 1 <= n
<= 10000. Each of the following n lines contains two integers a, b
separated by a single space, 0 <= a < b <= 10000. They are the
beginning and the end of an interval.

Output

Output the minimal number of elements in a set containing at least two different integers from each interval.

Sample Input

4
3 6
2 4
0 2
4 7

Sample Output

4

上一题差分约束的阉割版（传送门）

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <cstdlib>
 #include <algorithm>
 #include <queue>
 using namespace std;
 typedef long long ll;
 typedef pair<int,int> PII;
 #define MAXN 50010
 #define REP(A,X) for(int A=0;A<X;A++)
 #define INF  0x7fffffff
 #define CLR(A,X) memset(A,X,sizeof(A))
 struct node {
     int v,d,next;
 }edge[*MAXN];
 int head[MAXN];
 int e=;
 void init()
 {
     REP(i,MAXN)head[i]=-;
 }
 void add_edge(int u,int v,int d)
 {
     edge[e].v=v;
     edge[e].d=d;
     edge[e].next=head[u];
     head[u]=e;
     e++;
 }
 bool vis[MAXN];
 int dis[MAXN];
 void spfa(int s){
     CLR(vis,);
     REP(i,MAXN)dis[i]=i==s?:INF;
     queue<int>q;
     q.push(s);
     vis[s]=;
     while(!q.empty())
     {
         int x=q.front();
         q.pop();
         int t=head[x];
         while(t!=-)
         {
             int y=edge[t].v;
             int d=edge[t].d;
             t=edge[t].next;
             if(dis[y]>dis[x]+d)
             {
                 dis[y]=dis[x]+d;
                 if(vis[y])continue;
                 vis[y]=;
                 q.push(y);
             }
         }
         vis[x]=;
     }
 }
 int main()
 {
     ios::sync_with_stdio(false);
     int n;
     int u,v,d;
     int ans=;
     int maxn=;
     int minn=MAXN;
     while(scanf("%d",&n)!= EOF&&n)
     {
         e=;
         init();
         REP(i,n)
         {
             scanf("%d%d",&u,&v);
             add_edge(u,v+,-);
             maxn=max(maxn,v+);
             minn=min(u,minn);
         }
         for(int i=minn;i<maxn;i++){
             add_edge(i+,i,);
             add_edge(i,i+,);
         }
         spfa(minn);
         cout<<-dis[maxn]<<endl;
     }
     return ;
 }

代码君