Dining
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 10843 | Accepted: 4974 |
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
Line 1: Three space-separated integers: N, F, and D
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
Line 1: A single integer that is the maximum number of cows that can be fed both food and drink that conform to their wishes
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
One way to satisfy three cows is:
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.
题目大意：有F种食物，D种饮料，N头奶牛，只能吃某种食物和饮料(而且只能吃特定的一份)，一种食物被一头牛吃了之后，其余牛就不能吃了
第一行有N,F,D三个整数
接着2-N+1行代表第i头牛,前面两个整数是Fi与Di(食物与饮料的种类数量),接着是食物的种类与饮料的种类
要求输出最多分配能够满足的牛的数量
解题思路：建图,有2*n+f+d+2个顶点
0表示源点,2*n+f+d+1表示汇点
由源点指向食物,再由食物指向牛,牛再指向对应的饮料,饮料再指向汇点
当然要使每一头牛都对应每一份食物与饮料,所以应该牛i指向牛i再指向饮料,这样就可以避免一头牛只占用多份食物与饮料了
全部是有向的边,而且权值全部为1
我在这里是1到f为食物点，f+1到f+2*n为牛点，f+2*n+1到f+2*n+d为饮料点
代码：
type
edge=record
y,r,next,op:longint;
end;
var
g:array[0..10000] of edge;
level,q,h:array[0..10000] of longint;
n,m,f,i,j,a,b,c,tot,vs,ii,vt,x,y,k,ans:longint;
s:string;
function bfs:boolean;
var
i,f,r,tmp,v,u:longint;
begin
fillchar(level,sizeof(level),0);
f:=1; r:=1;
q[f]:=vs; level[vs]:=1;
repeat
v:=q[f]; tmp:=h[v];
while tmp<>-1 do
begin
u:=g[tmp].y;
if (g[tmp].r<>0) and (level[u]=0) then
begin
level[u]:=level[v]+1;
inc(r); q[r]:=u;
if u=vt then exit(true);
end;
tmp:=g[tmp].next;
end;
inc(f);
until f>r;
exit(false);
end;
function min(x,y:longint):longint;
begin
if x<y then
exit(x);
exit(y);
end;
function dfs(v,a:longint):longint;
var
ans,flow,tmp,u,value:longint;
begin
if (v=vt) or (a=0) then exit(a);
ans:=0; tmp:=h[v];
while tmp<>-1 do
begin
u:=g[tmp].y; value:=g[tmp].r;
if (level[u]=level[v]+1) then
begin
flow:=dfs(u,min(a,value));
if flow<>0 then
begin
g[tmp].r:=g[tmp].r-flow;
g[g[tmp].op].r:=g[g[tmp].op].r+flow;
ans:=ans+flow;
a:=a-flow;
if a=0 then break;
end;
end;
tmp:=g[tmp].next;
end;
exit(ans);
end;
procedure add(a,b,c:longint);
begin
inc(tot);
g[tot].y:=b;
g[tot].r:=c;
g[tot].next:=h[a];
h[a]:=tot;
g[tot].op:=tot+1;
inc(tot);
g[tot].y:=a;
g[tot].r:=0;
g[tot].next:=h[b];
h[b]:=tot;
g[tot].op:=tot-1;
end;
begin
while not eof do
begin
fillchar(g,sizeof(g),0);
fillchar(h,sizeof(h),$ff);
fillchar(level,sizeof(level),0);
fillchar(q,sizeof(q),0);
readln(n,f,m);
tot:=0;
ans:=0;
vs:=0;
vt:=2*n+f+m+1;
for i:=1 to n do
begin
x:=0;
y:=0;
a:=0;
add(i+f,i+n+f,1);
readln(s);
for j:=1 to length(s) do
begin
if s[j]=‘ ‘ then
break;
x:=x*10+ord(s[j])-ord(‘0‘);
end;
ii:=j+1;
for j:=ii to length(s) do
begin
if s[j]=‘ ‘ then
break;
y:=y*10+ord(s[j])-ord(‘0‘);
end;
k:=0;
ii:=j+1;
while k<x do
begin
if (s[ii]=‘ ‘) or (ii>length(s)) then
begin
add(a,f+i,1);
a:=0;
inc(ii);
inc(k);
end else
begin
a:=a*10+ord(s[ii])-ord(‘0‘);
inc(ii);
end;
end;
k:=0;
while k<y do
begin
if (s[ii]=‘ ‘) or (ii>length(s)) then
begin
add(f+n+i,f+2*n+a,1);
a:=0;
inc(ii);
inc(k);
end else
begin
a:=a*10+ord(s[ii])-ord(‘0‘);
inc(ii);
end;
end;
end;
for i:=1 to f do
add(0,i,1);
for i:=1 to m do
add(i+2*n+f,2*n+f+m+1,1);
while bfs do
begin
ans:=ans+dfs(vs,maxlongint);
end;
writeln(ans);
end;
end.
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