POJ3680 Intervals（最小费用最大流）

选择若干条线段使权值最大，并且点覆盖次数不超过k。

建图如下：vs到0建立容量为k费用为0的边；坐标终点到vt连接一条容量为k费用为0的边；对于每两个相邻坐标连接一条容量为INF费用为0的边；对于线段每两个端点连接一条容量1费用为-cost的边。

这样跑最小费用最大流。相当于找出k个线段集合，每个集合的线段都不重合。原问题就这样求解。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<queue>
 4 #include<algorithm>
 5 using namespace std;
 6 #define INF (1<<30)
 7 #define MAXN 444 
 8 #define MAXM 444*444*2 
 9 struct Edge{
10     int u,v,cap,cost,next;
11 }edge[MAXM];
12 int head[MAXN];
13 int NV,NE,vs,vt;
14 
15 void addEdge(int u,int v,int cap,int cost){
16     edge[NE].u=u; edge[NE].v=v; edge[NE].cap=cap; edge[NE].cost=cost;
17     edge[NE].next=head[u]; head[u]=NE++;
18     edge[NE].u=v; edge[NE].v=u; edge[NE].cap=0; edge[NE].cost=-cost;
19     edge[NE].next=head[v]; head[v]=NE++;
20 }
21 bool vis[MAXN];
22 int d[MAXN],pre[MAXN];
23 bool SPFA(){
24     for(int i=0;i<NV;++i){
25         vis[i]=0;
26         d[i]=INF;
27     }
28     vis[vs]=1;
29     d[vs]=0;
30     queue<int> que;
31     que.push(vs);
32     while(!que.empty()){
33         int u=que.front(); que.pop();
34         for(int i=head[u]; i!=-1; i=edge[i].next){
35             int v=edge[i].v;
36             if(edge[i].cap && d[v]>d[u]+edge[i].cost){
37                 d[v]=d[u]+edge[i].cost;
38                 pre[v]=i;
39                 if(!vis[v]){
40                     vis[v]=1;
41                     que.push(v);
42                 }
43             }
44         }
45         vis[u]=0;
46     }
47     return d[vt]!=INF;
48 }
49 int MCMF(){
50     int res=0;
51     while(SPFA()){
52         int flow=INF,cost=0;
53         for(int u=vt; u!=vs; u=edge[pre[u]].u){
54             flow=min(flow,edge[pre[u]].cap);
55         }
56         for(int u=vt; u!=vs; u=edge[pre[u]].u){
57             edge[pre[u]].cap-=flow;
58             edge[pre[u]^1].cap+=flow;
59             cost+=flow*edge[pre[u]].cost;
60         }
61         res+=cost;
62     }
63     return res;
64 }
65 
66 int from[222],to[222],cost[222];
67 int point[444],pn;
68 int main(){
69     int t,n,k;
70     scanf("%d",&t);
71     while(t--){
72         scanf("%d%d",&n,&k);
73         pn=0;
74         for(int i=0; i<n; ++i){
75             scanf("%d%d%d",from+i,to+i,cost+i);
76             point[pn++]=from[i]; point[pn++]=to[i];
77         }
78         sort(point,point+pn);
79         pn=unique(point,point+pn)-point;
80 
81         vs=pn; vt=vs+1; NV=vt+1; NE=0;
82         memset(head,-1,sizeof(head));
83 
84         addEdge(vs,0,k,0);
85         addEdge(pn-1,vt,k,0);
86         for(int i=1; i<pn; ++i){
87             addEdge(i-1,i,INF,0);
88         }
89         for(int i=0; i<n; ++i){
90             int x=lower_bound(point,point+pn,from[i])-point;
91             int y=lower_bound(point,point+pn,to[i])-point;
92             addEdge(x,y,1,-cost[i]);
93         }
94 
95         printf("%d
",-MCMF());
96     }
97     return 0;
98 }