poj 3281 最大流

poj 3281

题意：每只奶牛喜欢吃若干种食物和饮料，问只吃一种食物和饮料的奶牛最多有多少只。

思路：将奶牛做点拆开，形成源点-食物-奶牛（左）-奶牛（右）-饮料-汇点的链，每条边的容量都是1，奶牛（左）-奶牛（右）限制了一只奶牛同时最多吃一种食物喝一种饮料。

然后跑最大流就可以了。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;

const int MAXN = 1010;//点数的最大值
const int MAXM = 400010;//边数的最大值
const int INF = 0x3f3f3f3f;
struct Edge{
int to,next,cap,flow;
}edge[MAXM];//注意是MAXM
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];
void init(){
tol = 0;
memset(head,-1,sizeof(head));
}
//加边，单向图三个参数，双向图四个参数
void addedge(int u,int v,int w,int rw=0){
edge[tol].to = v;edge[tol].cap = w;edge[tol].next = head[u];
edge[tol].flow = 0;head[u] = tol++;
edge[tol].to = u;edge[tol].cap = rw;edge[tol].next = head[v];
edge[tol].flow = 0;head[v]=tol++;
}
int sap(int start,int ed,int N){
memset(gap,0,sizeof(gap));
memset(dep,0,sizeof(dep));
memcpy(cur,head,sizeof(head));
int u = start;
pre[u] = -1;
gap[0] = N;
int ans = 0;
while(dep[start] < N){
if(u == ed){
int Min = INF;
for(int i = pre[u];i != -1; i = pre[edge[i^1].to])
if(Min > edge[i].cap - edge[i].flow)
Min = edge[i].cap - edge[i].flow;
for(int i = pre[u];i != -1; i = pre[edge[i^1].to]){
edge[i].flow += Min;
edge[i^1].flow -= Min;
}
u = start;
ans += Min;
continue;
}
bool flag = false;
int v;
for(int i = cur[u]; i != -1;i = edge[i].next){
v = edge[i].to;
if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u]){
flag = true;
cur[u] = pre[v] = i;
break;
}
}
if(flag){
u = v;
continue;
}
int Min = N;
for(int i = head[u]; i != -1;i = edge[i].next)
if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min){
Min = dep[edge[i].to];
cur[u] = i;
}
gap[dep[u]]--;
if(!gap[dep[u]])return ans;
dep[u] = Min+1;
gap[dep[u]]++;
if(u != start) u = edge[pre[u]^1].to;
}
return ans;
}
int main() {
int n, f, d;
while(~scanf("%d %d %d", &n, &f, &d)) {
init();
for(int i = 1; i <= f; i++) addedge(1, n + n + i + 1, 1);
for(int i = 1; i <= d; i++) addedge(i + 1 + n + n + f, 2 + n + n + f + d, 1);
for(int i = 1; i <= n; i++) addedge(1 + i, 1 + n + i, 1);
for(int i = 1; i <= n; i++){
int fi, di;
scanf("%d %d", &fi, &di);
while(fi--){
int a;
scanf("%d", &a);
addedge(n + n + 1 + a, 1 + i, 1);
}
while(di--){
int a;
scanf("%d", &a);
addedge(1 + n + i, 1 + n + n + f + a, 1);
}
}
printf("%d\n", sap(1, n + n + f + d + 2, n + n + f + d + 2));
}
}