zoukankan      html  css  js  c++  java
  • 【 2018南京 】Magic Potion (网络流)

    There are nnn heroes and mmm monsters living in an island. The monsters became very vicious these days, so the heroes decided to diminish the monsters in the island. However, the iii-th hero can only kill one monster belonging to the set MiM_iMi. Joe, the strategist, has kkk bottles of magic potion, each of which can buff one hero's power and let him be able to kill one more monster. Since the potion is very powerful, a hero can only take at most one bottle of potion.

    Please help Joe find out the maximum number of monsters that can be killed by the heroes if he uses the optimal strategy.

    Input

    The first line contains three integers n,m,kn, m, kn,m,k (1≤n,m,k≤5001 le n, m, k le 5001n,m,k500) — ext{---}— the number of heroes, the number of monsters and the number of bottles of potion.

    Each of the next nnn lines contains one integer tit_iti, the size of MiM_iMi, and the following tit_{i}ti integers Mi,jM_{i, j}Mi,j (1≤j≤ti1 le j le t_i1jti), the indices (111-based) of monsters that can be killed by the iii-th hero (1≤ti≤m,1≤Mi,j≤m1 le t_ile m, 1leq M_{i, j} le m1tim,1Mi,jm).

    Output

    Print the maximum number of monsters that can be killed by the heroes.

    样例输入1

    3 5 2
    4 1 2 3 5
    2 2 5
    2 1 2

    样例输出1

    4

    样例输入2

    5 10 2
    2 3 10
    5 1 3 4 6 10
    5 3 4 6 8 9
    3 1 9 10
    5 1 3 6 7 10

    样例输出2

    7

    SOLUTION:


    本来的自身流量和额外k个流量是不能混在一起的
    因为有可能 别的hero的自身流量跑到了别的hero身上

    CODE:
    #include<iostream>
    #include<cstring>
    #include<cstdio>
    #include<algorithm>
    #include<queue>
    using namespace std;
    const int inf=1<<28;
    int cnt=1,head[130000];
    int n,m,s,t;
    inline int Read(){
        int x=0;
        char c=getchar();
        while(c>'9'||c<'0')c=getchar();
        while(c>='0'&&c<='9')x=x*10+c-'0',c=getchar();
        return x;
    }
    struct Node{
        int v;
        int next;
        int val;
    }node[300010];
    inline void addedge(int u,int v,int val){
        node[++cnt].v=v;
        node[cnt].val=val;
        node[cnt].next=head[u];
        head[u]=cnt;
    }
    int dep[130000],gap[130000];
    void bfs(){
        memset(dep,-1,sizeof(dep));
        memset(gap,0,sizeof(gap));
        dep[t]=0;
        gap[0]=1;
        queue<int>q;
        q.push(t);
        while(!q.empty()){
            int u=q.front();
            q.pop();
            for(int i=head[u];i;i=node[i].next){
                int v=node[i].v;
                if(dep[v]!=-1)continue;
                q.push(v);
                dep[v]=dep[u]+1;
                gap[dep[v]]++;
            }
        }
        return;
    }
    int maxflow;
    int dfs(int u,int flow){
        if(u==t){
            maxflow+=flow;
            return flow;
        }
        int used=0;
        for(int i=head[u];i;i=node[i].next){
            int d=node[i].v;
            if(node[i].val&&dep[d]+1==dep[u]){
                int mi=dfs(d,min(node[i].val,flow-used));
                if(mi){
                    node[i].val-=mi;
                    node[i^1].val+=mi;
                    used+=mi;
                }
                if(used==flow)return used;
            }
        }
        --gap[dep[u]];
        if(gap[dep[u]]==0)dep[s]=n+1;
        dep[u]++;
        gap[dep[u]]++;
        return used;
    }
    int ISAP(){
        maxflow=0;
        bfs();
        while(dep[s]<n)dfs(s,inf);
        return maxflow;
    }
    int k;
    int main(){
    
    
        n=Read(),m=Read(),k=Read();
       for(int  i=1;i<=n;i++)
       {
           int p;cin>>p;
           for(int j=1;j<=p;j++)
           {
               int x;x=Read();
               addedge(i,2*n+x,1);
               addedge(2*n+x,i,0);
               // cout<<i<<" "<<2*n+x<<endl;
           }
       }
    
       for(int i=1;i<=n;i++)
        addedge(n+i,i,2),addedge(i,i+n,0);
    
       for(int i=1;i<=m;i++)
        addedge(2*n+i,2*n+m+i,1),addedge(2*n+m+i,2*n+i,0);
    
       for(int i=1;i<=m;i++)
        addedge(2*n+m+i,2*n+2*m+3,inf),addedge(2*n+2*m+3,2*n+m+i,0);
    
       for(int i=1;i<=n;i++)
        addedge(2*n+2*m+2,n+i,1),addedge(n+i,2*n+2*m+2,0);
    
        for(int i=1;i<=n;i++)
            addedge(2*n+2*m+1,n+i,1),addedge(n+i,2*n+2*m+1,0);
    
       addedge(2*n+2*m+1,2*n+2*m+2,k);
       addedge(2*n+2*m+2,2*n+2*m+1,0);
    
        s=2*n+2*m+1,t=2*n+2*m+3;
        n=2*n+2*m+3;
    
       cout<<ISAP();
    }
    /*
    1 1 4
    1 1
    */
    

      










  • 相关阅读:
    每日Linux命令不完整命令
    mysql安装
    自动调用杀毒软件对文件进行杀毒
    正则获得字符串数组,以字符串分隔获取
    利用SQL语句查找某数据库中所有存储过程包含的内容
    TextBox的滚动条自动到最底部、利用枚举获取HashTable中的值
    Ajax中如何使用Session变量,Cookies可以用表单验证的方式获取并使用。
    用SQL语句批量生成一个表的INSERT语句
    利用DataSet、DataTable、DataView按照自定义条件过滤数据
    向线程传递数据与线程用回调方法检索数据
  • 原文地址:https://www.cnblogs.com/zhangbuang/p/11231662.html
Copyright © 2011-2022 走看看