poj 3281 Dining(网络流最大流+分点)

题目大意:有N种食物,和N种饮料.每头牛有喜欢的饮料和食物.问有多少头牛可以同时吃到自己喜欢的饮料和食物.

思路:这道题主要在于建图问题.

添加超级汇点和超级源点自然不用多说了.

正确的建模:源点-->food-->牛(左)-->牛(右)-->drink-->汇点
(转载的)

网上把这个叫做折点,不过我觉得交分点可能比较合适,使需要同时满足的条件都在一条路径上.

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define MAXN 2020
#define MAXM 8080
#define INF 0x7fffffff
int dep[MAXN], head[MAXN], gap[MAXN], cur[MAXN];
int cnt;
struct edge
{
int v;
int cap;
int flow;
int next;
}edg[MAXM];

void init()
{
cnt = 0;
memset(head, -1, sizeof(head));
}

void addedge(int u, int v, int w, int rw=0)
{
edg[cnt].v = v;
edg[cnt].cap = w;
edg[cnt].flow = 0;
edg[cnt].next = head[u];
head[u] = cnt++;
edg[cnt].v = u;
edg[cnt].cap = rw;
edg[cnt].flow = 0;
edg[cnt].next = head[v];
head[v] = cnt++;
}

int Q[MAXN];
void BFS(int start, int end)
{
memset(gap, 0, sizeof(gap));
memset(dep, -1, sizeof(dep));
gap[0] = 1;
dep[end] = 0;
int front =0,rear = 0;
Q[rear++] = end;
while (front != rear)
{
int u = Q[front++];
for (int i = head[u]; i != -1; i = edg[i].next)
{
int v = edg[i].v;
if (dep[v] != -1)continue;
dep[v] = dep[u] + 1;
gap[dep[v]]++;
Q[rear++] = v;
}
}
}

int S[MAXN];
int sap(int start, int end, int n)
{
memcpy(cur, head, sizeof(head));
BFS(start, end);
int top = 0;
int u = start;
int ans = 0;
while (dep[start] < n)
{
if (u == end)
{
int Min = INF;
int inser;
for (int i = 0; i < top; i++)
if (Min > edg[S[i]].cap - edg[S[i]].flow)
{
Min = edg[S[i]].cap - edg[S[i]].flow;
inser = i;
}
for (int i = 0; i < top; i++)
{
edg[S[i]].flow += Min;
edg[S[i] ^ 1].flow -= Min;
}
ans += Min;
top = inser;
u = edg[S[top] ^ 1].v;
continue;
}
bool flag = false;
int v;
for (int i = cur[u]; i != -1; i = edg[i].next)
{
v = edg[i].v;
if (edg[i].cap - edg[i].flow&&dep[v] + 1 == dep[u])
{
cur[u] = i;
flag = true;
break;
}
}
if (flag)
{
S[top++] = cur[u];
u = v;
continue;
}
int Min = n;
for (int i = head[u]; i != -1; i = edg[i].next)
{
if (edg[i].cap - edg[i].flow&&Min > dep[edg[i].v])
{
Min = dep[edg[i].v];
cur[u] = i;
}
}
gap[dep[u]]--;
if (!gap[dep[u]])return ans;
dep[u] = Min + 1;
gap[dep[u]]++;
if (u != start)u = edg[S[--top] ^ 1].v;
}
return ans;
}

int main()
{
int n, f, d;
while (~scanf("%d%d%d", &n, &f, &d))
{
init();
int E = 2*n + f + d + 1;
for (int i = 1; i <= f; i++)
addedge(0, i, 1);
for (int i = 1; i <= d; i++)
addedge(2*n + f + i, E, 1);
for (int i = 1; i <= n; i++)
addedge(f + i, f + i+n, 1);
int f1, d1,k;
for (int i = 1; i <= n; i++)
{
scanf("%d%d", &f1, &d1);
while (f1--)
{
scanf("%d", &k);
addedge(k, f + i, 1);
}
while (d1--)
{
scanf("%d", &k);
addedge(f + i + n, 2 * n + f + k, 1);
}
}
printf("%d\n", sap(0, E, E));
}
}