智算之道2020第三场初赛参考代码

题目

​​http://oj.csen.org.cn/contest/10?view=problems​​

T1

#include <bits/stdc++.h>

using namespace std;

int main()
{
    int n, L, A, B, temp, V;
    cin >> n >> L >> A >> B;

    for(int i = 0; i < n; i++)
    {
        int type, x;
        cin >> type;
        if(type == 1)
        {
            cin >> x;
            temp = x;
        }
        else if(type == 2)
        {
            cin >> x;
            V = x;
        }
        else if(type == 3)
        {
            if(temp >= A && temp <= B && V >= L)
            {
                cout << temp << endl;
            }
            else
            {
                cout << "GG" << endl;
            }
        }
    }
    return 0;
}

T2

#include <cstdio>

const int maxN = 4 * 100007;
int a[maxN], b[maxN];
int lastPosA[maxN], lastPosB[maxN];
bool aHas[maxN], bHas[maxN];


int main()
{
    //freopen("T2.txt", "r", stdin);


    int n;
    scanf("%d", &n);

    int cntA = 0, cntB = 0, posA = 1, posB = 1;

    //第一阶段，摸牌
    for(int i = 1, x; i <= 4 * n - 2; i++)
    {
        scanf("%d", &x);

        if(i % 2)
        {
            a[posA] = x;
            if(aHas[x])
            {
                cntA--;
            }
            else
            {
                cntA++;
                lastPosA[x] = posA;
            }

            posA++;
            aHas[x] = !aHas[x];
        }
        else
        {
            b[posB] = x;
            if(bHas[x])
            {
                cntB--;
            }
            else
            {
                cntB++;
                lastPosB[x] = posB;
            }

            posB++;
            bHas[x] = !bHas[x];
        }
    }

    //第二阶段，抽对方的牌
    bool aliceTurn = true; //Alice先抽
    int ans = 0;
    int j = 1, k = 1;
    while(cntA && cntB)
    {
        ans++;
        if(aliceTurn)//Alice抽Bob的牌
        {
            while(j < posB)
            {
                if(bHas[b[j]] && lastPosB[b[j]] == j)
                {
                    bHas[b[j]] = false; //b[j]被Alice抽走
                    cntB--;

                    if(aHas[b[j]])
                    {
                        cntA--;
                    }
                    else
                    {
                        cntA++;
                        lastPosA[b[j]] = posA;
                        a[posA++] = b[j];
                    }
                    aHas[b[j]] = !aHas[b[j]];
                    j++;
                    break;
                }
                else
                {
                    j++;
                }
            }//for
        }
        else //Bob抽Alice的牌
        {
            while(k < posA)
            {
                if(aHas[a[k]])
                {
                    aHas[a[k]] = false; //a[k]被Bob抽走
                    cntA--;

                    if(bHas[a[k]])
                    {
                        cntB--;
                    }
                    else
                    {
                        cntB++;
                        lastPosB[a[k]] = posB;
                        b[posB++] = a[k];
                    }

                    bHas[a[k]] = !bHas[a[k]];
                    k++;

                    break;
                }
                else
                {
                    k++;
                }
            }//while
        }

        aliceTurn = !aliceTurn;
    }//while

    printf("%d", ans);

    return 0;
}

T3

#include <cstdio>
#include <algorithm>
#include <vector>
#include <iostream>
using namespace std;

inline int read()
{
    int x=0;
    char t = getchar();
    while(t>'9'||t<'0')
    {
        t=getchar();
    }

    while(t>='0'&&t<='9')
    {
        x=x*10+t-'0';
        t=getchar();
    }

    return x;
}

const int N = 600000 + 10;
int n,k;
vector<int> edge[N];
int nodeCnt[N];

bool cmp(const int x, const int y)
{
    return nodeCnt[x] < nodeCnt[y];
}

void dfs(int u)
{
    nodeCnt[u] = 1;
    for(int i=0; i<edge[u].size(); i++)
    {
        int v = edge[u][i];
        dfs(v);
        nodeCnt[u] += nodeCnt[v];
    }

    sort(edge[u].begin(), edge[u].end(), cmp);
}

int main()
{
    freopen("T3.txt", "r", stdin);

    n = read();
    k = read();

    for(int v=2, fa; v<=n; v++)
    {
        fa = read(); //v节点的父节点
        edge[fa].push_back(v);
    }

    dfs(1);

    typedef long long LL;
    LL ans = 0;
    for(int u=1; u<=n; u++)
    {
        LL dist = 1;  //第一个子节点离父节点的距离是1
        for(int i=0; i<edge[u].size(); i++)
        {
            ans += dist;
            int v = edge[u][i];
            dist += nodeCnt[v]; //为计算下一个子节点离父节点的距离作准备
        }
    }

    printf("%lld\n", (ans+1)*k);

    return 0;
}