poj1422 Air Raid

本題若讀懂題目就會發現是一道最短路徑覆蓋的裸題，之後先把無向圖轉化爲二分圖，然後再用匈牙利算法求出其最大匹配數，最後最短路徑覆蓋數就是n-最大匹配數。

建圖方法：先拆點，將每個點分爲兩個點，左邊是1到n個點，右邊也是1到n個點,然後每一條有向邊對應左邊的點指向右邊的點即可

const MAXN=120;
var
	tc,n,m:longint;
	g:array[1..MAXN,1..MAXN]of longint;
	match,t:array[1..MAXN]of longint;
	h:array[1..MAXN]of boolean;
procedure init;
var i,x,y:longint;
begin
	fillchar(match,sizeof(match),0);
	fillchar(t,sizeof(t),0);
	read(n);
	read(m);
	for i:=1 to m do
	begin
		read(x,y);
		inc(t[x]);
		g[x,t[x]]:=y;
	end;
end;
function check(k:longint):boolean;
var i:longint;
begin
	for i:=1 to t[k] do
	if(not h[g[k,i]])then
	begin
		h[g[k,i]]:=true;
		if(match[g[k,i]]=0)or(check(match[g[k,i]]))then
		begin
			match[g[k,i]]:=k;
			exit(true);
		end;
	end;
	exit(false);
end;
procedure hungarian;
var i,ans:longint;
begin
	ans:=0;
	for i:=1 to n do
	begin
		fillchar(h,sizeof(h),0);
		if(check(i))then inc(ans);
	end;
	writeln(n-ans);
end;
begin
	readln(tc);
	while(tc<>0)do
	begin
		dec(tc);
		init;
		hungarian;
	end;
end.