Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 24379		Accepted: 9274

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. 

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.

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.

5
3 7 3
8 10 3
6 8 1
1 3 1
10 11 1

6

题意：给一个区间[ai,bi],这个区间里有ci个相同的元素,问这个集合最少包含多少元素;

思路：

可以得到这样的几个不等式;c[i]<=dis[b+1]-dis[a];0<=dis[i+1]-dis[i];-1<=dis[i]-dis[i+1];题意是一定要满足这些条件,那么就是找最长路;当然也可以变成>=来建图,求最短路;

AC代码：

//最长路径的代码//#include <bits/stdc++.h>
#include <iostream>
#include <queue>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
#define Riep(n) for(int i=1;i<=n;i++)
#define Riop(n) for(int i=0;i<n;i++)
#define Rjep(n) for(int j=1;j<=n;j++)
#define Rjop(n) for(int j=0;j<n;j++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=0x3f3f3f3f;
const int N=5e4+5;
int n,head[3*N],vis[N],ans[N],cnt,mmin;
struct Edge
{int to,next,dis;
}edge[3*N];
queue<int>qu;
void spfa()
{mst(vis,0);mst(ans,-inf);ans[mmin]=0;qu.push(mmin);vis[mmin]=1;while(!qu.empty()){int fr=qu.front();qu.pop();for(int i=head[fr];i!=-1;i=edge[i].next){int x=edge[i].to;if(ans[x]<ans[fr]+edge[i].dis){ans[x]=ans[fr]+edge[i].dis;if(!vis[x]){qu.push(x);vis[x]=1;}}}vis[fr]=0;}
}
void add_edge(int s,int e,int val)
{edge[cnt].to=e;edge[cnt].next=head[s];edge[cnt].dis=val;head[s]=cnt++;
}
int main()
{while(scanf("%d",&n)!=EOF){int a,b,c,mmax=0;mmin=55000;cnt=0;mst(head,-1);Riep(n){scanf("%d%d%d",&a,&b,&c);mmax=max(mmax,b);mmin=min(mmin,a);add_edge(a,b+1,c);}Riep(mmax)add_edge(i,i+1,0),add_edge(i+1,i,-1);spfa();printf("%d\n",ans[mmax+1]);}return 0;
}

//最短路径的代码
//#include <bits/stdc++.h>
#include <iostream>
#include <queue>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
#define Riep(n) for(int i=1;i<=n;i++)
#define Riop(n) for(int i=0;i<n;i++)
#define Rjep(n) for(int j=1;j<=n;j++)
#define Rjop(n) for(int j=0;j<n;j++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=0x3f3f3f3f;
const int N=5e4+5;
int n,head[3*N],vis[N],ans[N],cnt,mmin,mmax;
struct Edge
{int to,next,dis;
}edge[3*N];
queue<int>qu;
void spfa()
{mst(vis,0);mst(ans,inf);ans[mmax]=0;qu.push(mmax);vis[mmax]=1;while(!qu.empty()){int fr=qu.front();qu.pop();for(int i=head[fr];i!=-1;i=edge[i].next){int x=edge[i].to;if(ans[x]>ans[fr]+edge[i].dis){ans[x]=ans[fr]+edge[i].dis;if(!vis[x]){qu.push(x);vis[x]=1;}}}vis[fr]=0;}
}
void add_edge(int s,int e,int val)
{edge[cnt].to=e;edge[cnt].next=head[s];edge[cnt].dis=val;head[s]=cnt++;
}
int main()
{while(scanf("%d",&n)!=EOF){int a,b,c;mmax=0;mmin=55000;cnt=0;mst(head,-1);Riep(n){scanf("%d%d%d",&a,&b,&c);mmax=max(mmax,b+1);mmin=min(mmin,a);add_edge(b+1,a,-c);}Riep(mmax)add_edge(i,i+1,1),add_edge(i+1,i,0);spfa();printf("%d\n",-ans[mmin]);}return 0;
}