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=50010;
const int maxm=1000010;
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+1,inf);
    q[front=1]=S;back=2;
    vis[S]=1;
    dis[S]=0;
    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=0;
    while(~scanf("%d",&n)){
        memset(fir,0,sizeof(fir));
        cnt=0;maxl=0;        
        for(int i=1;i<=n;i++){
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            addedge(b,a-1,-c);
            maxl=max(maxl,b);
        }
        for(int i=1;i<=maxl;i++){
            addedge(i-1,i,1);
            addedge(i,i-1,0);
        }
        Spfa(maxl,0);
        printf("%d
",-dis[0]);
    }
}