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  • Evanyou Blog 彩带

      题目传送门

      这题的思路我觉得five20巨佬讲的已经非常清晰了,所以就推荐一下他的题解,我就只放代码了

    //It is made by HolseLee on 7th Feb 2018
    //Luogu.org P1251
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<cmath>
    #include<iostream>
    #include<queue>
    #include<algorithm>
    using namespace std;
    #define inf 1e12
    const int N=20001;
    typedef long long ll;
    ll n,d1,d2,c1,c2,p;
    ll sta,endd,head[N];
    ll cnt=1,ans,dis[N];
    bool vis[N],inq[N];
    struct Node{
      ll to,val;
      ll cost,next;
    }edge[N<<2];
    queue<ll>team;
    inline ll read()
    {
      char ch=getchar();ll num=0;bool flag=false;
      while(ch<'0'||ch>'9'){if(ch=='-')flag=true;ch=getchar();}
      while(ch>='0'&&ch<='9'){num=num*10+ch-'0';ch=getchar();}
      return flag?-num:num;
    }
    inline void add(ll x,ll y,ll z,ll c)
    {
      edge[++cnt].to=y;
      edge[cnt].val=z;
      edge[cnt].cost=c;
      edge[cnt].next=head[x];
      head[x]=cnt;
    }
    inline void add_edge(ll x,ll y,ll z,ll c)
    {add(x,y,z,c);add(y,x,0,-c);}
    inline bool spfa()
    {
      for(ll i=sta;i<=endd;i++)dis[i]=inf;
      memset(inq,false,sizeof(inq));
      team.push(endd);dis[endd]=0;
      inq[endd]=true;
      while(!team.empty()){
        ll u=team.front();team.pop();
        inq[u]=false;
        for(ll i=head[u];i!=-1;i=edge[i].next){
          ll v=edge[i].to;
          if(edge[i^1].val&&dis[v]>dis[u]-edge[i].cost){
        dis[v]=dis[u]-edge[i].cost;
        if(!inq[v]){
          inq[v]=true;
          team.push(v);
        }
          }
        }
      }
      return dis[sta]<inf;
    }
    inline ll dfs(ll u,ll nowflow)
    {
      vis[u]=true;
      if(nowflow==0||u==endd)return nowflow;
      ll used=0;
      for(ll i=head[u];i!=-1;i=edge[i].next){
        ll v=edge[i].to;
        if(!vis[v]&&edge[i].val&&dis[v]==dis[u]-edge[i].cost){
          ll ka=dfs(v,min(nowflow,edge[i].val));
          edge[i].val-=ka;edge[i^1].val+=ka;
          used+=ka;nowflow-=ka;
          if(nowflow==0)return used;
        }
      }
      return used;
    }
    void ready()
    {
      memset(head,-1,sizeof(head));
      n=read();ll x;
      sta=0;endd=2*n+1;
      for(ll i=1;i<=n;i++){
        x=read();
        add_edge(sta,i+n,x,0);
        add_edge(i,endd,x,0);}
      p=read();
      d1=read();c1=read();
      d2=read();c2=read();
      for(ll i=1;i<=n;i++){
        add_edge(sta,i,inf,p);
        if(i+d1<=n)add_edge(i+n,i+d1,inf,c1);
        if(i+d2<=n)add_edge(i+n,i+d2,inf,c2);
        if(i+1<=n)add_edge(i,i+1,inf,0);
      }
    }
    void work()
    {
      while(spfa()){
        vis[endd]=true;
        while(vis[endd]){
          memset(vis,false,sizeof(vis));
          ll ka=dfs(sta,inf);
          ans+=ka*dis[sta];
        }
      }
      printf("%lld",ans);
    }
    int main()
    {
      ready();
      work();
      return 0;
    }
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  • 原文地址:https://www.cnblogs.com/cytus/p/8427437.html
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