Intervals
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 18338 | Accepted: 6869 |
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.
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.
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
Source
思路：建边（a-1,b）<=-c;(a,a+1)<=-0;(a+1,a)<=1;
///1025MS
#include<iostream>
#include<cstring>
#include<queue>
#include<vector>
using namespace std;
const int mm=51010;
const int oo=1e9;
const int m=50004;
class node
{
public:int v,c,to;
}e[23*mm];
int q[mm];
int dis[mm],vis[mm],n,l,r,pre[mm],pos;
void add(int u,int v,int c)
{
e[pos].v=v;e[pos].c=c;e[pos].to=pre[u];pre[u]=pos++;
}
void spfa()
{ memset(vis,0,sizeof(vis));
for(int i=l;i<=r;i++)
dis[i]=oo;
dis[l]=0;
int s=0,t=1;
q[s]=l; vis[l]=1;int z;
while(s!=t)
{
z=q[s];s=(s+1)%m;vis[z]=0;
int v,c;
for(int p=pre[z];p!=-1;p=e[p].to)
{
v=e[p].v;c=e[p].c;
if(dis[v]>dis[z]+c)
{
dis[v]=dis[z]+c;
if(!vis[v])
{
vis[v]=1;q[t]=v;t=(t+1)%m;
}
}
}
}
}
int main()
{
while(cin>>n)
{ pos=0;
memset(pre,-1,sizeof(pre));
int a,b,c;l=oo;r=-oo;
memset(e,0,sizeof(e));
for(int i=0;i<n;i++)
{
cin>>a>>b>>c;
add(a-1,b,-c);
if(a<l)l=a;if(b>r)r=b;
}
--l;
///s[b]-s[a]<=c (a-1,b)中至少有c
for(int i=l;i<r;i++)
{ add(i+1,i,1);add(i,i+1,0);
}
///l^=r;r^=l;l^=r;
spfa();
cout<<dis[l]-dis[r]<<"\n";
}
}