Gym - 101981I-MagicPotion-最大流

题目描述

There are n heroes and m 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 i-th hero can only kill one monster belonging to the set Mi. Joe, the strategist, has k 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, k (1 ≤ n, m, k ≤ 500) — the number of heroes, the number of monsters and the number of bottles of potion.
Each of the next n lines contains one integer ti , the size of Mi, and the following ti integers Mi,j (1 ≤ j ≤ ti), the indices (1-based) of monsters that can be killed by the i-th hero (1 ≤ ti ≤ m, 1 ≤ Mi,j ≤ m).
Output
Print the maximum number of monsters that can be killed by the heroes.
Sample Input
sample input 1
3 5 2
4 1 2 3 5
2 2 5
2 1 2
sample input 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
Sample Output
sample output 1
4
sample output 2
7

题目分析

发现是匹配问题，思考将问题建图后用最大流算法进行解决。

如果没有k个药水，则很明确，将问题转化成一个二部图，从源点到英雄，从英雄到怪兽，从怪兽到汇点，路径上的容量都为1，则最大流就是杀死的最多的怪兽。

因为还有k个药水，并且每一个英雄都只能拿一个药水，我们从源点引出一个点，容量为k,再从这个点到每个英雄引一条路径，每条路径容量为1

AC代码

#include<iostream>
#include<cstring>
#include<cstdio>
#include<climits>
#include<algorithm>
#include<ctime>
#include<cstdlib>
#include<queue>
#include<set>
#include<map>
#include<cmath>using namespace std;typedef long long ll;
const int MAXN=1020;
const int INF=0x3f3f3f3f;struct Edge
{int from,to,cap,flow;Edge(int _from,int _to,int _cap,int _flow):from(_from),to(_to),cap(_cap),flow(_flow){}
};
struct Dinic
{int s,t;vector<Edge> edges;vector<int> G[MAXN];int d[MAXN]; bool vis[MAXN]; int cur[MAXN];void addEdge(int from,int to,int cap){edges.push_back((Edge){from,to,cap,0});edges.push_back((Edge){to,from,0,0});int tt=edges.size();G[from].push_back(tt-2);G[to].push_back(tt-1);}bool BFS(){memset(vis,0,sizeof(vis));queue<int>Q;Q.push(s);d[s]=0;vis[s]=1;while(!Q.empty()){int x=Q.front(); Q.pop();for(int i=0;i<G[x].size();++i){Edge &e =edges[G[x][i]];if(!vis[e.to] && e.cap>e.flow){vis[e.to]=1;d[e.to]=d[x]+1;Q.push(e.to);}}}return vis[t];}int DFS(int x,int a){if(x==t || a==0) return a;int flow=0,f;for(int&i=cur[x];i<G[x].size();++i){Edge& e=edges[G[x][i]];if(d[x]+1==d[e.to] && (f=DFS(e.to,min(a,e.cap-e.flow)))>0){e.flow+=f;edges[G[x][i]^1].flow-=f;flow+=f;a-=f;if(a==0) break;}}return flow;}int MF(int s,int t){this->s=s; this->t=t;int flow=0;while(BFS()){memset(cur,0,sizeof(cur));flow+=DFS(s,INF);}return flow;}
}dinic;int n,m,k;int main()
{scanf("%d%d%d",&n,&m,&k);int tmp=n+m+1;int t=tmp+1;int s=0;dinic.addEdge(s,tmp,k);for(int i=1;i<=n;i++){dinic.addEdge(s,i,1); dinic.addEdge(tmp,i,1);int t,u; scanf("%d",&t);while(t--){scanf("%d",&u); dinic.addEdge(i,u+n,1);}}for(int i=1;i<=m;i++){dinic.addEdge(i+n,t,1);}printf("%d\n",dinic.MF(s,t));return 0;
}