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[50010*3];
int head[50010];
int dis[50010];
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=0;
        cnt=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d%d",&l,&r,&k);
            addedge(l,r+1,k);
            maxx=max(r+1,maxx);
            minx=min(l,minx);
        }
        for(int i=minx;i<maxx;i++)
        {
            addedge(i,i+1,0);
            addedge(i+1,i,-1);
        };
        spfa(minx);
        printf("%d
",dis[maxx]);
    }
    return 0;
}
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]=0;
    Q.insert(s);
    int v,k;
    while(!Q.empty())
    {
        v=*Q.begin();
        Q.erase(v);
        k=head[v];
        while(k!=-1)
        {
            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 ;
}