图论(差分约束系统)：POJ 1201 Intervals

Intervals

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 24099		Accepted: 9159

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

　　

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #define inf 0x7ffffff
 using namespace std;
 const int maxn=;
 const int maxm=;
 int cnt,fir[maxn],to[maxm],val[maxm],nxt[maxm],dis[maxn],vis[maxn];
 void addedge(int a,int b,int v)
 {
     nxt[++cnt]=fir[a];
     to[cnt]=b;
     val[cnt]=v;
     fir[a]=cnt;
 }
 int q[maxm],front,back;
 void Spfa(int S,int T)
 {
     fill(dis,dis+S+,inf);
     q[front=]=S;back=;
     vis[S]=;
     dis[S]=;
     while(front<back)
     {
         int node=q[front++];vis[node]=false;
         for(int i=fir[node];i;i=nxt[i]){
             if(dis[to[i]]<=dis[node]+val[i])continue;
             dis[to[i]]=dis[node]+val[i];
             if(!vis[to[i]])
                 q[back++]=to[i];
             vis[to[i]]=true;
         }
     }
 }

 int main()
 {
     int n,maxl=;
     while(~scanf("%d",&n)){
         memset(fir,,sizeof(fir));
         cnt=;maxl=;
         for(int i=;i<=n;i++){
             int a,b,c;
             scanf("%d%d%d",&a,&b,&c);
             addedge(b,a-,-c);
             maxl=max(maxl,b);
         }
         for(int i=;i<=maxl;i++){
             addedge(i-,i,);
             addedge(i,i-,);
         }
         Spfa(maxl,);
         printf("%d\n",-dis[]);
     }
 }