# 2018 ACM-ICPC Asia Nanjing Regional Contest I. Magic Potion （最大流）

## Problem I. Magic Potion

• Input file: standard input
• Output file: standard output

### Description

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.

### Examples

• standard input

3 5 2
4 1 2 3 5
2 2 5
2 1 2
• standard output

4
• standard input

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
• standard output

7

### 链接

https://codeforces.com/gym/101981

### 题意

n个人、m个怪兽和k瓶膜法药水，正常情况下每个人可以杀死一个怪兽，现让人服用药水，每个人至多一瓶药水，可以多杀一个怪物，问最多能杀死多少个怪兽。

### 题解 ### 代码

• Accepted 31 ms 12100 KB
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
const int maxn = 1010;

int n, m, k;

struct Edge
{
int u, v, cap;
Edge () {}
Edge (int _u, int _v, int _cap): u(_u), v(_v), cap(_cap) {}
}edge[maxn*maxn];

int R, S, T;

vector <int> E[maxn];
int dis[maxn];
int current[maxn];

void addedge(int u, int v, int cap)
{
E[u].push_back(R);
edge[R++] = Edge(u, v, cap);
E[v].push_back(R);
edge[R++] = Edge(v, u, 0);
}

int bfs()
{
memset(dis, 0x3f, sizeof dis);
queue <int> Q;
Q.push(S);
dis[S] = 0;
while (!Q.empty())
{
int u = Q.front();
Q.pop();
for (int i = 0; i < E[u].size(); ++i)
{
Edge &e = edge[E[u][i]];
if (e.cap > 0 && dis[e.v] == 0x3f3f3f3f)
{
dis[e.v] = dis[u] + 1;
Q.push(e.v);
}
}
}
return dis[T] < 0x3f3f3f3f;
}

int dinic(int u, int maxflow)
{
if (u == T)
return maxflow;
for (int i = current[u]; i < E[u].size(); ++i)
{
current[u] = i;
Edge &e = edge[E[u][i]];
if (dis[e.v] == dis[u] + 1 && e.cap > 0)
{
int flow = dinic(e.v, min(maxflow, e.cap));
if (flow)
{
e.cap -= flow;
edge[E[u][i]^1].cap += flow;
return flow;
}
}
}
return 0;
}

int DINIC()
{
int ans = 0;
while (bfs())
{
int flow;
memset(current, 0, sizeof current);
while (flow = dinic(S, 0x3f3f3f3f))
ans += flow;
}
return ans;
}

int main()
{
int t, u;
scanf("%d%d%d", &n, &m, &k);
R = 0;
S = 0, T = n + m + 2;
for (int i = 0; i <= T; ++i)
E[i].clear();
memset(edge, 0, sizeof edge);
for (int i = 1; i <= n; ++i) // 人与怪物之间连边
{
scanf("%d", &t);
for (int j = 1; j <= t; ++j)
{
scanf("%d", &u);
}
}
for (int i = 1; i <= n; ++i) // 起点与人之间连边
for (int i = 1; i <= m; ++i) // 怪物与终点之间连边
}