POJ 1201 Intervals（差分约束）

思路：和POJ1716基本一样，只是这里区间不同元素的个数也给了而已

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
#define inf -1e9
const int maxn = 50005;
vector<pair<int,int> >e[maxn];
int d[maxn],cnt[maxn],inq[maxn],n;
int spfa(int num)
{
memset(cnt,0,sizeof(cnt));
memset(inq,0,sizeof(inq));
for(int i = 0;i<=num;i++)
d[i]=inf;
inq[0]=1;
queue<int>q;
d[0]=0;
q.push(0);
while(!q.empty())
{
int u = q.front();
q.pop();
inq[u]=0;
for(int i =0;i<e[u].size();i++)
{
int v = e[u][i].first;
if(d[v]<d[u]+e[u][i].second)
{
d[v]=d[u]+e[u][i].second;
if(!inq[v])
{
inq[v]=1;
q.push(v);
if(++cnt[v]>n)
return false;
}
}
}
}
return 1;
}
int main()
{
int num = 0;
while(scanf("%d",&n)!=EOF)
{
for(int i = 0;i<=num;i++)
e[i].clear();
num = 0;
for(int i = 1;i<=n;i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
u++,v++;
num = max(num,v);
e[u-1].push_back(make_pair(v,w));
}
for(int i = 1;i<=num;i++)
{
e[0].push_back(make_pair(i,0));
e[i].push_back(make_pair(i+1,0));
e[i+1].push_back(make_pair(i,-1));
}
spfa(num);
printf("%d\n",d[num]);
}
}


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