HDU 1054 Strategic Game 二分图最小点覆盖

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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。

http://acm.hdu.edu.cn/showproblem.php?pid=1054

/*********************************************    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-8using 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;}

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