hdu 1384 Intervals (差分约束）

Intervals

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4181    Accepted Submission(s): 1577

Problem 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 endpoints 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 <= 50 000) - 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 <= 50 000 and 1 <= ci <= bi - ai + 1.

Process to the end of file.

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

Author

1384

 

---

一道很明显的差分约束题。

对于每个[l,r]对应的k可以列出方程 f[r]-f[l-1]>=k;

同时区间之间 f[i]-f[i-1]>=0; f[i-1]-f[i]>=-1;

然后就是拿所有区间的最左端点作源点跑一遍spfa（）;

结束。

代码：

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<vector>
 #include<set>
 #define INF 1000000000
 #define clr(x) memset(x,0,sizeof(x))
 #define clrmin(x) memset(x,-1,sizeof(x))
 #define clrmax(x) memset(x,0x3f3f3f3f,sizeof(x))
 using namespace std;
 struct node
 {
     int to,val,next;
 }edge[*];
 int head[];
 int dis[];
 int n,maxx,minx,maxans,cnt,l,r,k;
 set<int> Q;
 void addedge(int l,int r,int k);
 int min(int a,int b);
 int max(int a,int b);
 void spfa(int s);
 int main()
 {
     while(scanf("%d",&n)!=EOF)
     {
         clrmin(head);
         clrmin(dis);
         Q.clear();
         maxx=;
         cnt=;
         for(int i=;i<=n;i++)
         {
             scanf("%d%d%d",&l,&r,&k);
             addedge(l,r+,k);
             maxx=max(r+,maxx);
             minx=min(l,minx);
         }
         for(int i=minx;i<maxx;i++)
         {
             addedge(i,i+,);
             addedge(i+,i,-);
         };
         spfa(minx);
         printf("%d\n",dis[maxx]);
     }
     return ;
 }
 void addedge(int l,int r,int k)
 {
     edge[++cnt].to=r;
     edge[cnt].val=k;
     edge[cnt].next=head[l];
     head[l]=cnt;
     return;
 }
 int min(int a,int b)
 {
     return a<b?a:b;
 }
 int max(int a,int b)
 {
     return a>b?a:b;
 }
 void spfa(int s)
 {
     dis[s]=;
     Q.insert(s);
     int v,k;
     while(!Q.empty())
     {
         v=*Q.begin();
         Q.erase(v);
         k=head[v];
         while(k!=-)
         {
             if(dis[v]+edge[k].val>dis[edge[k].to])
             {
                 dis[edge[k].to]=dis[v]+edge[k].val;
                 Q.insert(edge[k].to);
             }
             k=edge[k].next;
         }
     }
     return ;
 }