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  • 网络流(最大费用最大流) :POJ 3680 Intervals

    Intervals
    Time Limit: 5000MS   Memory Limit: 65536K
    Total Submissions: 7218   Accepted: 3011

    Description

    You are given N weighted open intervals. The ith interval covers (ai, bi) and weighs wi. 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 ≤ KN ≤ 200).
    The next N line each contain three integers ai, bi, wi(1 ≤ ai < bi ≤ 100,000, 1 ≤ wi ≤ 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
      
      这题就是求最大区间K覆盖,最大费用最大流走起~~~
      1 #include <algorithm> 
      2 #include <iostream>
      3 #include <cstring>
      4 #include <cstdio>
      5 #include <queue>
      6 using namespace std;
      7 const int INF=233333333;
      8 const int maxn=2010,maxm=20010;
      9 int cnt,fir[maxn],nxt[maxm],to[maxm],cap[maxm],val[maxm],dis[maxn],path[maxn];
     10 struct data{
     11     int l,r,v;
     12     bool operator <(const data A)const{
     13         if(l!=A.l)
     14             return l<A.l;
     15         return r<A.r;     
     16     }
     17 }ar[maxn];
     18 
     19 struct data2{
     20     int a,pos;
     21     bool operator <(const data2 A)const{
     22         if(a!=A.a)
     23             return a<A.a;
     24         return pos<A.pos;     
     25     }
     26 }br[maxn];
     27 
     28 void addedge(int a,int b,int c,int v)
     29 {
     30     nxt[++cnt]=fir[a];to[cnt]=b;cap[cnt]=c;val[cnt]=v;fir[a]=cnt;
     31 }
     32 int S,T;
     33 int Spfa()
     34 {
     35     queue<int>q;
     36     memset(dis,-1,sizeof(dis));
     37     q.push(S);dis[S]=0;
     38     while(!q.empty())
     39     {
     40         int node=q.front();q.pop();
     41         for(int i=fir[node];i;i=nxt[i])
     42             if(cap[i]&&dis[node]+val[i]>dis[to[i]]){
     43                 dis[to[i]]=val[i]+dis[node];
     44                 path[to[i]]=i;
     45                 q.push(to[i]);
     46             }
     47     }
     48     return dis[T]==-1?0:dis[T]; 
     49 }
     50 
     51 int Aug()
     52 {
     53     int p=T,f=INF;
     54     while(p!=S)
     55     {
     56         f=min(f,cap[path[p]]);
     57         p=to[path[p]^1];
     58     }
     59     p=T;
     60     while(p!=S)
     61     {
     62         cap[path[p]]-=f;
     63         cap[path[p]^1]+=f;
     64         p=to[path[p]^1];
     65     }
     66     return f;
     67 }
     68 
     69 int MCMF()
     70 {
     71     int ret=0,d;
     72     while(d=Spfa())
     73         ret+=Aug()*d;
     74     return ret;    
     75 }
     76 
     77 int cont;
     78 void Init()
     79 {
     80     cnt=1;cont=0;
     81     memset(fir,0,sizeof(fir));
     82 }
     83 
     84 int main()
     85 {
     86     int Q,n,k;
     87     scanf("%d",&Q);
     88     while(Q--)
     89     {
     90         Init();
     91         scanf("%d%d",&n,&k);    
     92         for(int i=1;i<=n;i++)
     93             scanf("%d%d%d",&ar[i].l,&ar[i].r,&ar[i].v);
     94         sort(ar+1,ar+n+1);
     95         for(int i=1;i<=n;i++){
     96             br[i*2-1].a=ar[i].l;br[i*2].a=ar[i].r;
     97             br[i*2-1].pos=br[i*2].pos=i;
     98         }
     99         sort(br+1,br+2*n+1);
    100         for(int i=1;i<=n;i++)
    101             ar[i].l=ar[i].r=0;
    102         for(int i=1;i<=2*n;i++)
    103         {
    104             int p=br[i].pos;
    105             if(!ar[p].l){
    106                 ar[p].l=++cont;
    107             }
    108             else{
    109                 ar[p].r=++cont;
    110             }
    111         }
    112 
    113         S=0;T=cont+1;
    114         for(int i=0;i<cont+1;i++)
    115             addedge(i,i+1,k,0),addedge(i+1,i,0,0); 
    116         for(int i=1;i<=n;i++)
    117             addedge(ar[i].l,ar[i].r,1,ar[i].v),addedge(ar[i].r,ar[i].l,0,-ar[i].v);
    118         printf("%d
    ",MCMF());        
    119     }
    120 
    121     return 0;
    122 }
    尽最大的努力,做最好的自己!
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  • 原文地址:https://www.cnblogs.com/TenderRun/p/5225437.html
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