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
题意:
给出数轴上的n个闭合int型区间。现在要在数轴上任意取一堆元素,构成一个元素集合V,要求给出的每个区间和元素集合V的交集至少有两个不同的元素,求集合V最小的元素个数。
根据题目的信息,我们可以得出d[y]-d[x-1]>=c;
然后我们转换一下:
d[y]>=d[x-1]+c;
不要忘了一下条件:
d[i]-d[i-1]>=0;
d[i]-d[i-1]<=0;
没有这些条件边与边之间是串不起来的。
然后再求最长路就可以了。
代码如下:
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
#include <vector>
using namespace std;
const int maxn=50005;
const int INF=0x3f3f3f3f;
int n;
struct edge
{
int e,len;
};
vector <edge> ve[maxn];
int d[maxn];
int maxx,minn;
int vis[maxn];
void init()
{
maxx=-1;
minn=INF;
for (int i=0;i<maxn;i++)
d[i]=-INF;
for (int i=0;i<maxn;i++)
ve[i].clear();
memset (vis,0,sizeof(vis));
}
void spfa()
{
d[minn]=0;
queue<int>q;
q.push(minn);
while (!q.empty())
{
int u=q.front(); q.pop();
vis[u]=0;
for (int i=0;i<ve[u].size();i++)
{
int v=ve[u][i].e;
if(d[v]<d[u]+ve[u][i].len)
{
d[v]=d[u]+ve[u][i].len;
if(!vis[v])
{
q.push(v);
vis[v]=1;
}
}
}
}
printf("%d\n",d[maxx]);
}
int main()
{
scanf("%d",&n);
init();
for (int i=0;i<n;i++)
{
int x,y,len;
scanf("%d%d%d",&x,&y,&len);
maxx=max(maxx,y);
minn=min(minn,x-1);
edge temp;
temp.e=y; temp.len=len;
ve[x-1].push_back(temp);
}
for (int i=minn+1;i<=maxx;i++)
{
edge temp1,temp2;
temp1.e=i; temp1.len=0;
temp2.e=i-1; temp2.len=-1;
ve[i-1].push_back(temp1);
ve[i].push_back(temp2);
}
spfa();
return 0;
}