poj3680 最大权不相交路径

Intervals

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 8587		Accepted: 3662

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

14
12
100000
100301

题意：给你N个区间段的(a,b)和价值，让你在不相交的情况下求m次求得最大值。

题解：先将每个点都离散化，然后依次添加INF的边，然后输入m条边，用lower_bound找到x,y相应下标，然后加边权值为这个区间段的相反数，跑最小费用流就可以了



题解这里

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int N=500;
const int M=1e4+88;
const int INF=0x3f3f3f3f;
int head[N],tot,pre[N],C[N],F[N],V[N],n,m;
struct node{
   int u,v,flow,cost,next;
}e[M];
void add(int u,int v,int flow,int cost){
   e[tot].u=u;e[tot].v=v;e[tot].flow=flow;e[tot].cost=cost;e[tot].next=head[u];head[u]=tot++;
   e[tot].u=v;e[tot].v=u;e[tot].flow=0;e[tot].cost=-cost;e[tot].next=head[v];head[v]=tot++;
}
int SPFA(int s,int t){
   memset(pre,-1,sizeof(pre));
   for(int i=1;i<=t+1;++i) F[i]=0,C[i]=INF,V[i]=0;
   queue<int>Q;
   Q.push(s);
   C[0]=0,F[0]=INF,V[0]=1;
   while(!Q.empty()){
     int u=Q.front();
     Q.pop();
     V[u]=0;
     for(int i=head[u];~i;i=e[i].next){
        int v=e[i].v,f=e[i].flow,c=e[i].cost;
        if(f>0&&C[v]>C[u]+c) {
            C[v]=C[u]+c;
            pre[v]=i;
            F[v]=min(f,F[u]);
            if(!V[v]) V[v]=1,Q.push(v);
        }
     }
   }
   return F[t]&&C[t]!=0;
}
int MCMF(int s,int t){
   int ans=0,temp;
   while(temp=SPFA(s,t)){
    for(int i=pre[t];~i;i=pre[e[i].u]) {
        ans+=temp*e[i].cost;
        e[i].flow-=temp;
        e[i^1].flow+=temp;
    }
   }
   return ans;
}
struct point{
   int x,y,val;
}Po[N];
int ar[N];
int main(){
   int T;
   for(scanf("%d",&T);T--;){
    scanf("%d%d",&n,&m);
    tot=0;
    memset(head,-1,sizeof(head));
    int ct=0;
    for(int i=1;i<=n;++i) {scanf("%d%d%d",&Po[i].x,&Po[i].y,&Po[i].val);
    ar[++ct]=Po[i].x,ar[++ct]=Po[i].y;
    }
    sort(ar+1,ar+ct+1);
    int num=2;
    for(int i=2;i<=ct;++i) {
        while(ar[i]==ar[num-1]&&i<ct) ++i;
        if(i<=ct) ar[num++]=ar[i];
    }
    for(int i=2;i<=num;++i) add(i-1,i,INF,0);
    add(0,1,m,0);
    add(num,num+1,m,0);
    for(int i=1;i<=n;++i) {
        int l=lower_bound(ar+1,ar+num+1,Po[i].x)-ar;
        int r=lower_bound(ar+1,ar+num+1,Po[i].y)-ar;
        add(l,r,1,-Po[i].val);
    }
    printf("%d\n",-MCMF(0,num+1));
   }
}