[HDOJ6178]Monkeys（二分图最大匹配）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6178

题意：希望在一棵树中删掉最多的边，使得剩下的k个点至少有一个点相连。

比赛时候学弟贪心dfs过的，然后发现这个题实际上是最大匹配。

任意两点相连代表匹配中的一条边，最大匹配后的结果就是树上最多有多少点对两两有边。

考虑最大匹配Max>=k的时候，直接两两配对即可，答案就是k/2取上整。

Max<k的时候就是k-Max/2取下整。

朝鲜也流行卡常啊？我夸夸你们啊？

#include <bits/stdc++.h>
using namespace std;

typedef struct Edge { int v, next; }Edge;
const int maxn = 100005;
const int inf = 0x3f3f3f3f;
int nx, ny, dist;
int Mx[maxn], My[maxn], dx[maxn], dy[maxn], vis[maxn];
int head[maxn], ecnt;
Edge edge[maxn<<1];

int n, m;
void init() {
    memset(head, -1, sizeof(head));
    for(int i = 0; i < maxn; i++) edge[i].next = -1;
    ecnt = 0;
}

void adde(int u, int v) {
    edge[ecnt].v = v;
    edge[ecnt].next = head[u]; head[u] = ecnt++;
}

bool Dfs(int u) {
    for(int i = head[u]; ~i; i=edge[i].next) {
        int& v = edge[i].v;
        if(!vis[v] && dy[v] == dx[u] + 1) {
            vis[v] = 1;
            if(My[v] != -1 && dy[v] == dist) continue;
            if(My[v] == -1 || Dfs(My[v])) {
                My[v] = u;
                Mx[u] = v;
                return 1;
            }
        }
    }
    return 0;
}
bool Search() {
    queue<int> Q;
    dist = inf;
    memset(dx, -1, sizeof(dx));
    memset(dy, -1, sizeof(dy));
    for(int i = 0; i < nx; i++) {
        if(Mx[i] == -1) {
            Q.push(i), dx[i] = 0;
        }
    }
    while(!Q.empty()) {
        int u = Q.front();
        if(dx[u] > dist) break;
        Q.pop();
        for(int i = head[u]; ~i; i=edge[i].next) {
            int& v = edge[i].v;
            if(dy[v] == -1) {
                dy[v] = dx[u] + 1;
                if(My[v] == -1) dist = dy[v];
                else {
                    dx[My[v]] = dy[v] + 1;
                    Q.push(My[v]); 
                }
            }
        }
    }
    return dist != inf;
}

int MaxMatch() {
    int res = 0;
    memset(Mx, -1, sizeof(Mx));
    memset(My, -1, sizeof(My));
    while(Search()) {
        memset(vis, 0, sizeof(vis));
        for(int i = 0; i < nx; i++) {
            if(Mx[i] == -1 && Dfs(i)) res++;
        }
    }
    if(res >= m) return (m + 1) / 2;
    else return m - res / 2;
}

namespace fastIO {
    #define BUF_SIZE 100000
    //fread -> read
    bool IOerror = 0;
    inline char nc() {
        static char buf[BUF_SIZE], *p1 = buf + BUF_SIZE, *pend = buf + BUF_SIZE;
        if (p1 == pend) {
            p1 = buf;
            pend = buf + fread(buf, 1, BUF_SIZE, stdin);
            if (pend == p1) {
                IOerror = 1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch == ' ' || ch == '
' || ch == '
' || ch == '	';
    }
    inline void read(int &x) {
        char ch;
        while (blank(ch = nc()));
        if (IOerror)
            return;
        for (x = ch - '0'; (ch = nc()) >= '0' && ch <= '9'; x = x * 10 + ch - '0');
    }
    #undef BUF_SIZE
};


signed main() {
    // freopen("in", "r", stdin);
    int T, u;
    fastIO::read(T);
    while(T--) {
        init();
        fastIO::read(n); fastIO::read(m);
        nx = ny = n;
        for(int i = 2; i <= n; i++) {
            fastIO::read(u);
            adde(u-1, i-1), adde(i-1, u-1);
        }
        printf("%d
", MaxMatch());
    }
    return 0;
}