POJ 3680 Intervals

最新推荐文章于 2020-01-16 20:09:33 发布

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#poj #网络流 #最小费用最大流

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本文介绍了一种使用网络流算法解决区间覆盖问题的方法，旨在选取一系列权重最高的开放区间，同时确保每个点被覆盖次数不超过限定值。通过离散化处理和网络流优化策略实现了高效的求解。

Intervals

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 7424		Accepted: 3100

Description

You are given N weighted open intervals. The ith interval covers (a_i, b_i) and weighs w_i. Your task is to pick some of the intervals to maximize the total weights under the limit that no point in the real axis is covered more than k times.

Input

The first line of input is the number of test case.
The first line of each test case contains two integers, N and K (1 ≤ K ≤ N ≤ 200).
The next N line each contain three integers a_i, b_i, w_i(1 ≤ a_i < b_i ≤ 100,000, 1 ≤ w_i ≤ 100,000) describing the intervals.
There is a blank line before each test case.

Output

For each test case output the maximum total weights in a separate line.

Sample Input

4

3 1
1 2 2
2 3 4
3 4 8

3 1
1 3 2
2 3 4
3 4 8

3 1
1 100000 100000
1 2 3
100 200 300

3 2
1 100000 100000
1 150 301
100 200 300

Sample Output

Source

【题目分析】

一看到巨大的ai，bi和超级小的n，k就知道有必要进行离散化（就是因为离散化出错，RE了好多遍，有可能并没有重复，需要特判一下）。然后就不会做了，关键是每一个点覆盖不能超过k次，用网络流的方法确实是很巧妙的，加入（S，1，k，0）（T-1 ,T ,k ,0 ）（ai，bi，1,-wi）(i,i+1,k,0)如果一个区间被选择，那么这个区间可能通过的流量就减去了1（S点出来的流量为k，那么最大流就是k）然后就轻而易举的解决了。

【代码】

#include <cstdio>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
using namespace std;
const int inf =0x3f3f3f3f;
const int maxn=10010;
const int maxm=10010;
#define M(a) memset(a,-1,sizeof a)
#define M0(a) memset(a,0,sizeof a)
#define MB(a) memset(a,0x3f,sizeof a)
int n,m,S,T,ans,en=0;
int v[maxn],u[maxn],cost[maxn],f[maxn],h[maxn],ne[maxn];
bool vis[maxm];//u->v
int l[maxn],r[maxn],w[maxn],ss[maxn];
int dis[maxm],with[maxm],minn[maxm];
inline void add(int a,int b,int r,int c)
{
	u[en]=a;v[en]=b;ne[en]=h[a];f[en]=r;cost[en]=c; h[a]=en++;
	u[en]=b;v[en]=a;ne[en]=h[b];f[en]=0;cost[en]=-c;h[b]=en++;
}
inline bool tell()
{
queue<int>q;
	M0(vis);MB(dis);MB(minn);M0(with);
	q.push(S);vis[S]=true;dis[S]=0;
	while (!q.empty())
	{
		int x=q.front();q.pop();vis[x]=false;
		for (int i=h[x];i!=-1;i=ne[i])
		{
			int y=v[i];
			if (dis[y]>dis[x]+cost[i]&&f[i]>0)
			{
				dis[y]=dis[x]+cost[i];
				minn[y]=min(minn[x],f[i]);
				with[y]=i;
				if (!vis[y]) vis[y]=1,q.push(y);
			}
		}
	}
	if (dis[T]==0x3f3f3f3f) return false;
	return true;
}
inline int zeng()
{
	for (int i=T;i!=S;i=v[with[i]^1])
	{
		f[with[i]]-=minn[T];
		f[with[i]^1]+=minn[T];
	}
	return minn[T]*dis[T];
}
inline void solve()
{
	M(v);M(u);M(h);M(ne);
	int n,k;
	scanf("%d%d",&n,&k);
	S=0;en=0;ans=0;
	for (int i=1;i<=n;++i){
		scanf("%d%d%d",&l[i],&r[i],&w[i]);
		ss[i*2-1]=l[i];
		ss[i*2]=r[i];
	}
	sort(ss+1,ss+2*n+1);
	for (int i=2*n;i>1;--i) if (ss[i]==ss[i-1]) ss[i]=inf;
	sort(ss+1,ss+2*n+1);
	int flag=0;
	for (int i=1;i<=2*n;i++)
	{
		if (ss[i]==inf) {T=i+1;flag=1;break;}
	}
	if (!flag) T=2*n+1;
	for (int i=1;i<T;++i)add(i,i+1,k,0);
	for (int i=1;i<=n;++i){
		for (int j=1;j<T;++j) if (l[i]==ss[j]) {l[i]=j;break;}
		for (int j=1;j<T;++j) if (r[i]==ss[j]) {r[i]=j;break;}
	}
	for (int i=1;i<=n;++i){
		add(l[i],r[i],1,-w[i]);
	}
	add(S,1,k,0);
	add(T-1,T,k,0);
	while (tell()) ans-=zeng();
	printf("%d\n",ans);
}
int main()
{
	int tt=0;
	scanf("%d",&tt);
	for (int z=1;z<=tt;++z) solve();
}