洛谷P1231
参考博客大佬Siyuan
题意
有n1本书,n2本练习册,n3本答案。给出m1个书和练习册的联系,m2个书和答案的联系,一本书一本练习册一本答案为一册,问最多可以组成多少册。
思路
如果只有两样东西就是典型的最大二分匹配问题,但这里有三种,所以匈牙利算法做不了,可以用最大流。建图方法是:源点->练习册->书(拆点)->答案->汇点。拆点很重要,下面是一个样例说明为什么要拆点,拆点也就是一个点拆成两个的意思。
如果没有拆点,下面的图用最大流跑出来就是2,显然答案并不是2,因为数只有一本。
将书的点拆成两个点,两点之间连一条容量为1的边,这样就可以限制书只用一次的条件了,答案得出就是1。
#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 4e4 + 7;
typedef long long ll;
struct Edge {
int from, to, cap, flow;
Edge(int u, int v, int c, int f):from(u), to(v), cap(c), flow(f) {}
};
struct Dinic {
int n, m, s, t;
vector<Edge> edges;
vector<int> G[maxn];
int d[maxn], cur[maxn];
bool vis[maxn];
void init(int n) {
for (int i = 0; i < n; i++) G[i].clear();
edges.clear();
}
void add(int from, int to,int cap) {
edges.push_back(Edge(from, to, cap, 0));
edges.push_back(Edge(to, from, 0, 0));
m = edges.size();
G[from].push_back(m - 2);
G[to].push_back(m - 1);
}
bool BFS() {
memset(vis, 0, sizeof(vis));
queue<int> que;
que.push(s);
d[s] = 0;
vis[s] = 1;
while (!que.empty()) {
int x = que.front();
que.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;
que.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 Maxflow(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;
}
};
int main()
{
int n, m, s, t, q;
Dinic solve;
scanf("%d%d%d", &n, &m, &q);
s = 0; t = n * 2 + m + q + 1;//m n n q
for (int i = 1; i <= n; i++) solve.add(m + i, m + n + i, 1);
for (int i = 1; i <= m; i++) solve.add(s, i, 1);
for (int i = 1; i <= q; i++) solve.add(2 * n + m + i, t, 1);
int m1, m2;
scanf("%d", &m1);
for (int i = 1; i <= m1; i++) {
int x, y;
scanf("%d%d", &x, &y);
solve.add(y, x + m, 1);
}
scanf("%d", &m2);
for (int i = 1; i <= m2; i++) {
int x, y;
scanf("%d%d", &x, &y);
solve.add(x + m + n, y + m + n + n, 1);
}
printf("%d\n", solve.Maxflow(s, t));
}