| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 10755 | Accepted: 4930 |
Description
Cows are such finicky eaters. Each cow has a preference for certain foods and drinks, and she will consume no others.
Farmer John has cooked fabulous meals for his cows, but he forgot to check his menu against their preferences. Although he might not be able to stuff everybody, he wants to give a complete meal of both food and drink to as many cows as possible.
Farmer John has cooked F (1 ≤ F ≤ 100) types of foods and prepared D (1 ≤ D ≤ 100) types of drinks. Each of his N (1 ≤ N ≤ 100) cows has decided whether she is willing to eat a particular food or drink a particular drink. Farmer John must assign a food type and a drink type to each cow to maximize the number of cows who get both.
Each dish or drink can only be consumed by one cow (i.e., once food type 2 is assigned to a cow, no other cow can be assigned food type 2).
Input
Lines 2.. N+1: Each line i starts with a two integers Fi and Di, the number of dishes that cow i likes and the number of drinks that cow i likes. The next Fi integers denote the dishes that cow i will eat, and the Di integers following that denote the drinks that cow i will drink.
Output
Sample Input
4 3 3 2 2 1 2 3 1 2 2 2 3 1 2 2 2 1 3 1 2 2 1 1 3 3
Sample Output
3
Hint
Cow 1: no meal
Cow 2: Food #2, Drink #2
Cow 3: Food #1, Drink #1
Cow 4: Food #3, Drink #3
The pigeon-hole principle tells us we can do no better since there are only three kinds of food or drink. Other test data sets are more challenging, of course.
题目意思：有n个牛，f种食物，d种饮料（食物和饮料只能选用一次）。每个牛都有自己喜爱的食物和饮料，问最大能使多少头牛吃到自己喜爱的食物和饮料。
显然这道题目 匹配。。。 我不会。那么只能最大流了。。。
错误思路：建立超级源点和超级汇点。让源点指向食物，食物指向牛，牛再指向饮料，最后让饮料指向汇点，中间所有边权为1。求最大流。
为什么错误？
错误在这一步：食物指向牛，牛再指向饮料。因为这样不能保证一头牛最多得到一种食物和一种饮料（有可能贪吃。。。）
正确思路：建立超级源点和超级汇点，把每头牛拆分成左，右牛。让源点指向食物，食物指向左牛，左牛指向右牛，右牛再指向饮料，最后让饮料指向汇点，所有边权为1。求最大流。
当然改动点时看个人爱好，我的是左牛为1到n，右牛n+1到2*n，食物2*n+1到2*n+f，饮料2*n+f+1到2*n+f+d。超级源为0，超级汇为2*n+f+d+1。
dinic算法：32ms
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>
#define MAXN 400+10//食物从2*n+1开始 到 2*n+f 饮料从2*n+f+1 开始 到2*n+1+f+d
#define MAXM 200000+100
#define INF 200000+10
using namespace std;
struct Edge
{
int from, to, cap, flow, next;
}edge[MAXM];
int dist[MAXN], vis[MAXN], head[MAXN], cur[MAXN], top;
int n, f, d;
void init()
{
top = 0;
memset(head, -1, sizeof(head));
}
void addedge(int u, int v, int w)
{
Edge E1 = {u, v, w, 0, head[u]};
edge[top] = E1;
head[u] = top++;
Edge E2 = {v, u, 0, 0, head[v]};
edge[top] = E2;
head[v] = top++;
}
void getmap()
{
int i, j;
int fnum, dnum;
for(i = 1; i <= n; i++)
{
scanf("%d%d", &fnum, &dnum);
while(fnum--)//食物 编号2*n+j
{
scanf("%d", &j);
addedge(2*n+j, i, 1);//食物和对应奶牛连通
}
while(dnum--)//饮料 编号2*n+f+j
{
scanf("%d", &j);
addedge(n+i, 2*n+f+j, 1);//奶牛和对应饮料连通
}
addedge(i, n+i, 1);//奶牛自己和自己连一次
}
for(i = 1; i <= f; i++)
addedge(0, 2*n+i, 1);//超级源点连通所有食物
for(i = 1; i <= d; i++)
addedge(2*n+f+i, 2*n+f+d+1, 1);//所有饮料连通超级汇点
}
bool BFS(int start, int end)
{
queue<int> Q;
memset(dist, -1, sizeof(dist));
memset(vis, 0, sizeof(vis));
while(!Q.empty()) Q.pop();
Q.push(start);
dist[start] = 0;
vis[start] = 1;
while(!Q.empty())
{
int u = Q.front();
Q.pop();
for(int i = head[u]; i != -1; i = edge[i].next)
{
Edge E = edge[i];
if(!vis[E.to] && E.cap > E.flow)
{
vis[E.to] = 1;
dist[E.to] = dist[u] + 1;
if(E.to == end) return true;
Q.push(E.to);
}
}
}
return false;
}
int DFS(int x, int a, int end)
{
if(x == end || a == 0) return a;
int flow = 0, f;
for(int& i = cur[x]; i != -1; i = edge[i].next)
{
Edge& E = edge[i];
if(dist[E.to] == dist[x]+1 && (f = DFS(E.to, min(a, E.cap-E.flow), end)) > 0)
{
E.flow += f;
edge[i^1].flow -= f;
flow += f;
a -= f;
if(a == 0) break;
}
}
return flow;
}
int Maxflow(int start, int end)
{
int flow = 0;
while(BFS(start, end))
{
memcpy(cur, head, sizeof(head));
flow += DFS(start, INF, end);
}
return flow;
}
int main()
{
while(scanf("%d%d%d", &n, &f, &d) != EOF)
{
init();
getmap();
printf("%d\n", Maxflow(0, 2*n+f+d+1));
}
return 0;
}