POJ 3680 Intervals 费用流构图题（区间的一定考虑相邻的）-淘宝拒签后怎么处理-程序博客网

Description

You are given N weighted open intervals. Theith interval covers (a_i,b_i) and weighsw_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 andK (1 ≤K ≤N ≤ 200).
The next N line each contain three integersa_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

43 11 2 22 3 43 4 83 11 3 22 3 43 4 83 11 100000 1000001 2 3100 200 3003 21 100000 1000001 150 301100 200 300

Sample Output

1412100000100301

题意：给定 N 个带权的开区间，第 i 个区间覆盖(ai, bi)，权为 wi。现在要你挑出一些区间使得总权值最大，并且满足实轴上任意一个点被覆盖不超过 K 次，每个区间选一次。注意区间内的点是实数。（1 <= K <= N<= 200, 1 <= ai < bi <= 100,000, 1 <= wi <= 100,000）

第一点：离散化处理。

第二点：求最大值，保存负数。

第三点：每个区间只能选择一次，所以对于区间（a，b），我们add(a,b,1,-w)

第四点：对于相邻的数字都要建立。（i，i+1，k，0）；

涉及到区间的一定要考虑相邻数字的连接问题

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<queue>
#include<set>
#include<cmath>
#include<bitset>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson i<<1,l,mid
#define rson i<<1|1,mid+1,r
#define llson j<<1,l,mid
#define rrson j<<1|1,mid+1,r
#define INF 0x7fffffff
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
#define maxn 20005
struct
{
int v,w,c,next,re;
//re记录逆边的下标,c是费用,w是流量
} e[maxn];
int n,cnt;
int head[maxn],que[maxn*8],pre[maxn],dis[maxn];
bool vis[maxn];
void add(int u, int v, int w, int c)
{
e[cnt].v=v,e[cnt].w=w,e[cnt].c=c;
e[cnt].next=head[u];
e[cnt].re=cnt+1,head[u]=cnt++;
e[cnt].v=u,e[cnt].w=0,e[cnt].c=-c;
e[cnt].next=head[v];
e[cnt].re=cnt-1,head[v]=cnt++;
}
bool spfa()
{
int i, l = 0, r = 1;
for(i = 0; i <= n; i ++)
dis[i] = INF,vis[i] = false;
dis[0]=0,que[0]=0,vis[0]=true;
while(l<r)
{
int u=que[l++];
for(i=head[u];i!=-1;i=e[i].next)
{
int v = e[i].v;
if(e[i].w&&dis[v]>dis[u]+e[i].c)
{
dis[v] = dis[u] + e[i].c;
pre[v] = i;
if(!vis[v])
{
vis[v] = true;
que[r++] = v;
}
}
}
vis[u] = false;
}
return dis[n]!=INF;
}
int change()
{
int i,p,sum=INF,ans=0;
for(i=n;i!=0;i=e[e[p].re].v)
{//e[e[p].re].v为前向结点,不理解看add和spfa
p=pre[i];//p为前向结点编号
sum=min(sum,e[p].w);
}
for(i=n;i!=0;i=e[e[p].re].v)
{
p=pre[i];
e[p].w-=sum;
e[e[p].re].w+=sum;
ans+=sum*e[p].c;//c记录的为单位流量费用，必须得乘以流量。
}
return ans;
}
int EK()
{
int sum=0;
while(spfa()) sum+=change();
return sum;
}
void init()
{
mem(head,-1),mem(pre,0),cnt=0;
}
struct abc
{
int u,v,w;
}aa[405];
int main()
{
//freopen("1.txt","r",stdin);
int t;
scanf("%d",&t);
while(t--)
{
init();
int N,K,i,j,x,y,tmp=0,a[402];
map<int,int>Map;
scanf("%d%d",&N,&K);
for(i=0;i<N;i++)
{
scanf("%d%d%d",&aa[i].u,&aa[i].v,&aa[i].w);
a[tmp++]=aa[i].u,a[tmp++]=aa[i].v;
}
sort(a,a+tmp);
int k=unique(a,a+tmp)-a;
for(i=0,tmp=0;i<k;i++)
Map[a[i]]=++tmp;
for(i=0;i<N;i++)
{
int u=Map[aa[i].u],v=Map[aa[i].v];
add(u,v,1,-aa[i].w);
}
for(i=1;i<tmp;i++)
add(i,i+1,K,0);
n=tmp+1;
add(0,1,K,0),add(tmp,n,K,0);
printf("%d\n",-EK());
}
return 0;
}