HDU 1054 Strategic Game 二分图最小点覆盖

题意：给你一颗树，一个点可以占领跟它连的所有的边。问最少需要多少个这样的点？

思路：一眼就是染色之后二分图最小点覆盖。

坑点：T得我不能自理，发现是边数组开小了，我加的双向边，也是爽。

/*********************************************
Problem : HDU 1054
Author  : NMfloat
InkTime (c) NM . All Rights Reserved .
********************************************/

#include <map>
#include <set>
#include <queue>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>

#define rep(i,a,b)  for(int i = a ; i <= b ; i ++)
#define rrep(i,a,b) for(int i = b ; i >= a ; i --)
#define repE(p,u) for(Edge * p = G[u].first ; p ; p = p -> next)
#define cls(a,x)   memset(a,x,sizeof(a))
#define eps 1e-8

using namespace std;

const int MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const int MAXN = 2000;
const int MAXE = 4000;

typedef long long LL;
typedef unsigned long long ULL;

struct Edge { //记录边
int to;
Edge * next;
}E[MAXE],E1[MAXE],*EE;

struct Gragh { //记录图的结点
Edge * first;
}G[MAXN],G1[MAXN];

int N,M;//二分图左右结点的个数
bool visit[MAXN];
int match[MAXN];//v2中匹配的情况
int C[MAXN];
int X[MAXN];//重新编号。
int Y[MAXN];

void addedge(int u,int v) { //加边
EE->to = v ; EE -> next = G[u].first ; G[u].first = EE ++;
//EE->to = u ; EE -> next = G[v].first ; G[v].first = EE ++;
}

void addedge1(int u,int v) { //加边
EE->to = v ; EE -> next = G1[u].first ; G1[u].first = EE ++;
EE->to = u ; EE -> next = G1[v].first ; G1[v].first = EE ++;
}

int T,n,m,k;

void init() {
EE = E1; N = M = 0;
cls(G,0);
cls(G1,0);
}

bool find_path(int u) {
int v;
repE(p,u) {
v = p->to;
if(!visit[v]) {
visit[v] = 1;
if(match[v] == -1 || find_path(match[v])) {//v没有匹配或者v可以找到另一条路径
match[v] = u;
return true;
}
}
}
return false;
}

int Max_match() {
cls(match,-1);
int cnt = 0;
rep(i,1,N) {
cls(visit,0);
if(find_path(i)) cnt ++;
}
return cnt;
}

void dfs(int u,int color,int fa = 0) {
C[u] = color % 2;
if(C[u] == 0) X[u] = ++N;//重新编号。
else Y[u] = ++M;
for(Edge * p = G1[u].first;p;p=p->next) {
int v = p->to;
if(v != fa) dfs(v,color+1,u);
}
}

void create_G(int u,int fa = 0) {
for(Edge * p = G1[u].first;p;p=p->next) {
int v = p->to;
if(v != fa)
{
if(C[u] == 0) addedge(X[u],Y[v]+N);
else addedge(X[v],Y[u]+N);
create_G(v,u);
}
}
}

void input() {
int fa , num , son;
rep(i,1,n) {
scanf("%d:(%d)",&fa,&num);
rep(i,1,num) {
scanf("%d",&son);
addedge1(fa,son);
}
}
}

void solve() {
//染色。0是X部;
dfs(0,0);
//rep(i,0,n-1) printf("%d ",C[i]);
//printf("%d\n",N);
EE = E;
create_G(0);
printf("%d\n",Max_match());
}

int main(void) {
//freopen("a.in","r",stdin);
while(~scanf("%d",&n)) {
init();
input();
solve();
}
return 0;
}