Intervals
Time Limit: 2000MS
Memory Limit: 65536K
Total Submissions: 22307
Accepted: 8413
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.
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
Source
Southwestern Europe 2002
题目大意：给你N个整数点构成的区间[ai，bi](ai，bi都为整数)，在区间[ai，bi]上最少选ci个点。
ci可在区间[ai，bi]中随意取，但是不能重复。问：要满足在N个区间取点，至少要选多少个点。
思路：差分约束思想。设Si为前i项的整数个数，则S(bi) - S(ai-1) >= ci。还有两个隐含约束条件
S(i-1) - S(i) <= 0，S(i)-S(i-1) <= 1。把这三种约束构建一个差分约束系统，用SPFA求最短路径。
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int MAXN = 50050;
const int INF = 0xffffff0;
struct EdgeNode
{
int to;
int w;
int next;
}Edges[MAXN << 2];
int Head[MAXN],Dist[MAXN],vis[MAXN];
int SPFA(int Left,int Right)
{
for(int i = Left; i <= Right+1; ++i)
Dist[i] = INF;
memset(vis,0,sizeof(vis));
queue<int> Q;
Q.push(Right);
vis[Right] = 1;
Dist[Right] = 0;
while( !Q.empty() )
{
int u = Q.front();
Q.pop();
vis[u] = 0;
for(int i = Head[u]; i != -1; i = Edges[i].next)
{
int temp = Dist[u] + Edges[i].w;
if( temp < Dist[Edges[i].to])
{
Dist[Edges[i].to] = temp;
if( !vis[Edges[i].to] )
{
vis[Edges[i].to] = 1;
Q.push(Edges[i].to);
}
}
}
}
return -Dist[Left];
}
int main()
{
int N,Left,Right,a,b,c;
while(~scanf("%d",&N))
{
Left = MAXN,Right = 0;
memset(Head,-1,sizeof(Head));
memset(Edges,0,sizeof(Edges));
int id = 0;
for(int i = 0; i < N; ++i)
{
scanf("%d%d%d",&a,&b,&c);
Edges[id].to = a;
Edges[id].w = -c;
Edges[id].next = Head[b+1];
Head[b+1] = id++;
if(a < Left)
Left = a;
if(b > Right)
Right = b;
}
Right++;
for(int i = Left; i < Right; ++i)
{
Edges[id].to = i;
Edges[id].w = 0;
Edges[id].next = Head[i+1];
Head[i+1] = id++;
Edges[id].to = i+1;
Edges[id].w = 1;
Edges[id].next = Head[i];
Head[i] = id++;
}
printf("%d\n",SPFA(Left,Right));
}
return 0;
}