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： ai bi ci 代表序列在[ai,bi]之间最少ci个整数，问你这个序列最少有多少个整数

题解：由差分约束系统，首先可以得到的是 dis[b+1] - dis[a] >=c[i] 

然后 这样还不够建图 对于每一个点 还有以下的不等式

dis[b+1]-dis[b] >=0

dis[b]-dis[b+1]>=-1

这样就可以建图了

题目的意思就可以转化为 从minm 到maxm最少有多少个整数了

#include <bits/stdc++.h>
#define maxn 500000+5
#define INF 0x3f3f3f3f
using namespace std;
int n,dis[maxn],head[maxn],len;
int vis[maxn],maxm=-INF,minm=INF;
struct edge{
    int to,val,next;
}e[maxn];
void add(int from ,int to ,int val){
    e[len].to=to;
    e[len].val=val;
    e[len].next=head[from];
    head[from]=len++;
}
void spfa(int s){
    memset(vis,0,sizeof(vis));
    for(int i=minm;i<=maxm;i++){
        dis[i]=-INF;
    }
    queue<int>q;
    q.push(s);
    vis[s]=1;
    dis[s]=0;
    while(!q.empty()){
       // cout<<233;
        int cur =q.front();q.pop();
        vis[cur]=0;
        for(int i=head[cur];i!=-1;i=e[i].next){
            int id=e[i].to;
            if(dis[id]<dis[cur]+e[i].val){
                dis[id]=dis[cur]+e[i].val;
                if(!vis[id]){
                    vis[id]=1;
                    q.push(id);
                }
            }
        }
    }
}
int main(){
    while(scanf("%d",&n)!=EOF){
        len=0;
        memset(head,-1,sizeof(head));
        for(int i=0;i<n;i++){
            int u,v,c;
            scanf("%d%d%d",&u,&v,&c);
            minm=min(u,minm);
            maxm=max(v+1,maxm);
            add(u,v+1,c);
        }
        for(int i=minm;i<maxm;i++){
            add(i,i+1,0);
            add(i+1,i,-1);
        }
      // cout<<233;
        spfa(minm);
        printf("%d\n",dis[maxm]);
    }
    return 0;
}

（poj崩了。。）