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  • POJ-3680:Intervals (费用流)

    You are given N weighted open intervals. The ith interval covers (aibi) 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 ≤ K ≤ N ≤ 200).
    The next N line each contain three integers aibiwi(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

    题意:给定N个带权线段,现在选一些线段,其和最大,而且每个点不被超过K个点覆盖。

    思路:离散化,费用流模板题,求最大费用最大流,最大输出-ans。横向i->i+1,加边(i,i+1,K,0);对于线段,加边(u,v,1,-cost);

    (N个线段满足u<v,所以不用考虑成环的问题。

    #include<deque>
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<iostream>
    #include<algorithm>
    #define ll long long
    #define maxn 410
    #define inf 1<<30
    using namespace std;
    int To[maxn*10],Laxt[maxn*10],Next[maxn*10],cap[maxn*10],cost[maxn*10],dis[maxn*10];
    int N,S,T,cnt,tot,ans; //费用
    bool inq[maxn],vis[maxn];
    deque<int>q;
    void init() { cnt=1; ans=0; memset(Laxt,0,sizeof(Laxt)); S=0; T=tot+1; }
    void add(int u,int v,int c,int cc) { Next[++cnt]=Laxt[u];Laxt[u]=cnt;To[cnt]=v;cap[cnt]=c;cost[cnt]=cc; }
    bool spfa()
    {
        memset(inq,0,sizeof(inq));
        for(int i=0;i<=T;i++) dis[i]=inf;
        inq[T]=1; dis[T]=0; q.push_back(T);
        while(!q.empty())
        {    
            int u=q.front(); q.pop_front();
            inq[u]=0;
            for(int i=Laxt[u];i;i=Next[i])
            {
                int v=To[i];
                if(cap[i^1]&&dis[v]>dis[u]-cost[i])
                {
                    dis[v]=dis[u]-cost[i];
                    if(!inq[u]){
                        inq[v]=1;
                        if(q.empty()||dis[v]>dis[q.front()]) q.push_back(v);
                        else q.push_front(v);
                    }
                }
            }
        }
        return dis[S]<inf;
    }
    int dfs(int u,int flow)
    {
        vis[u]=1;
        if(u==T||flow==0) return flow;
        int tmp,delta=0;
        for(int i=Laxt[u];i;i=Next[i])
        {
            int v=To[i];
            if((!vis[v])&&cap[i]&&dis[v]==dis[u]-cost[i])
            {
                tmp=dfs(v,min(cap[i],flow-delta));
                delta+=tmp; cap[i]-=tmp; cap[i^1]+=tmp;
            }
        }
        return delta;
    }
    int main()
    {
        int Case,N,K,i,j;
        scanf("%d",&Case);
        while(Case--){
            scanf("%d%d",&N,&K);
            int u[210],v[210],w[210],b[420]; tot=0;
            for(i=1;i<=N;i++) {
                scanf("%d%d%d",&u[i],&v[i],&w[i]);
                b[++tot]=u[i]; b[++tot]=v[i];
            }
            sort(b+1,b+tot+1);
            tot=unique(b+1,b+tot+1)-(b+1);
            init();
            for(i=1;i<=N;i++){
                u[i]=lower_bound(b+1,b+tot+1,u[i])-b;
                v[i]=lower_bound(b+1,b+tot+1,v[i])-b;
                add(u[i],v[i],1,-w[i]); add(v[i],u[i],0,w[i]);
            }
            add(S,1,K,0);  add(1,S,0,0);
            add(T-1,T,K,0); add(T,T-1,0,0);
            for(i=1;i<T;i++) add(i,i+1,K,0),add(i+1,i,0,0);
            while(spfa())
            {
               vis[T]=1;
               while(vis[T])
               {
                   memset(vis,0,sizeof(vis));
                   int tmp=dfs(S,inf);
                   ans+=(ll)tmp*dis[S];
               }
            }
            printf("%d
    ",-ans);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/hua-dong/p/9391556.html
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