HDU-2017 多校训练赛10-1008-Monkeys

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ACM模版

描述

题解

这个题主要是卡 I/O，需要一个输入外挂，让人头痛的是普通的输入外挂还不行，需要一个十分强大的外挂……当然，这个外挂我一会儿要添加到我的模版中，真的不错。

代码

#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <queue>using namespace std;const int MAXN = 1e5 + 7;int n, k;int pd[MAXN];int vis[MAXN];vector<int> arr[MAXN];struct FastIO{    static const int S = 100 << 1;    int wpos;    char wbuf[S];    FastIO() : wpos(0) {}    inline int xchar()    {        static char buf[S];        static int len = 0, pos = 0;        if (pos == len)        {            pos = 0;            len = (int)fread(buf, 1, S, stdin);        }        if (pos == len)        {            return -1;        }        return buf[pos++];    }    inline int xint()    {        int s = 1, c = xchar(), x = 0;        while (c <= 32)        {            c = xchar();        }        if (c == '-')        {            s = -1;            c = xchar();        }        for (; '0' <= c && c <= '9'; c = xchar())        {            x = x * 10 + c - '0';        }        return x * s;    }    ~FastIO()    {        if (wpos)        {            fwrite(wbuf, 1, wpos, stdout);            wpos = 0;        }    }} io;int main(){//    freopen("/Users/zyj/Desktop/input.txt", "r", stdin);    memset(pd, 0, sizeof(pd));    int T;    scanf("%d", &T);    while (T--)    {        memset(vis, 0, sizeof(vis));        n = io.xint();        k = io.xint();        for (int i = 1; i <= n; i++)        {            arr[i].clear();        }        int a;        for (int i = 2; i <= n; i++)        {            a = io.xint();            arr[a].push_back(i);            arr[i].push_back(a);            pd[a]++;            pd[i]++;        }        queue<int> q;        for (int i = 1; i <= n; i++)        {            if (pd[i] == 1)            {                q.push(i);            }            pd[i] = 0;        }        int ans = 0;        while (!q.empty())        {            int now = q.front();            q.pop();            if (arr[now].size() != 0)            {                int v = arr[now][0];                if (vis[now] == 0 && vis[v] == 0)                {                    ans++;                    vis[now] = 1;                    vis[v] = 1;                }                arr[v].erase(remove(arr[v].begin(), arr[v].end(), now),arr[v].end());                if (arr[v].size() == 1)                {                    q.push(v);                }            }        }        if (ans * 2 >= k)        {            printf("%d\n", (k + 1) / 2);        }        else        {            printf("%d\n", ans + k - ans * 2);        }    }    return 0;}

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