D. Playoff Tournament

合集 - 树论(34)

1.P3478 [POI2008] STA-Station2024-03-192.E1. Weights Division (easy version)2024-03-223.P5007 DDOSvoid 的疑惑2024-03-254.P2168 [NOI2015] 荷马史诗2024-03-305.P8306 【模板】字典树2024-03-306.P1481 魔族密码2024-03-307.P3128 [USACO15DEC] Max Flow P2024-04-028.P3258 [JLOI2014] 松鼠的新家2024-04-029.P5536 【XR-3】核心城市2024-04-0210.P5836 [USACO19DEC] Milk Visits S2024-04-0211.P3384 【模板】重链剖分/树链剖分2024-04-0312.P3038 [USACO11DEC] Grass Planting G2024-04-0313.P4551 最长异或路径2024-04-0414.F. Alex's whims2024-04-1215.J. 上学2024-05-1416.Game on Tree2024-05-1417.E. We Need More Bosses2024-05-1518.A. Party2024-05-1519.D. Strong Vertices2024-05-1620.B. Omkar and Heavenly Tree2024-05-1621.B. Mahmoud and Ehab and the bipartiteness2024-05-1622.P8655 [蓝桥杯 2017 国 B] 发现环2024-05-2423.P10298 [CCC 2024 S4] Painting Roads2024-05-2424.F. Gardening Friends2024-06-14

25.D. Playoff Tournament2024-06-19

26.E. Lomsat gelral2024-07-1227.F. Minimum Maximum Distance2024-07-1428.D. Book of Evil2024-07-1529.P2195 HXY造公园2024-07-1530.P3304 [SDOI2013] 直径2024-07-1531.D. Sasha and a Walk in the City2024-07-1932.D. Vitaly and Cycle2024-08-0133.P10838 『FLA - I』庭中有奇树2024-08-0734.F - Perfect Matching on a Tree2024-08-10

收起

原题链接

题解

1.观察样例易得改变是有后继性的
2.总是解题时没有思路，一点拨就有思路，怎么办？
我们要找到一些规律，即一定会发生的现象
3.当我们对一场比赛的晋级规律进行修改后，这场比赛的晋级人数取决于其先前两场比赛的晋级人数

实施

1.参考线段树，将第 $i$ 场比赛改为id为 $2^k-i$ 这样就能像线段树一样修改了
2.如果有问号，这场比赛至少晋升2人

code

#include<bits/stdc++.h>
using namespace std;
int a[1<<19]={0};
int id[1<<19]={0};
int main()
{
    int k;
    cin>>k;

    string s;
    cin>>s;
    s=' '+s;

    int n=(1<<k);
    for(int i=1;i<n;i++)
    {
        if(s[i]=='?')
        {
            a[n-i]=max(a[(n-i)*2]+a[(n-i)*2+1],2);
            id[n-i]=2;
        }
        else if(s[i]=='1')
        {
            a[n-i]=max(1,a[(n-i)*2]);
            id[n-i]=1;
        }
        else
        {
            a[n-i]=max(1,a[(n-i)*2+1]);
            id[n-i]=0;
        }
    }
    int q;
    cin>>q;
    while(q--)
    {
        int p;
        char c;
        cin>>p>>c;

        p=n-p;
        if(c=='1') id[p]=1;
        else if(c=='?') id[p]=2;
        else id[p]=0;

        while(p)
        {
            if(id[p]==2) a[p]=max(2,a[p*2]+a[p*2+1]);
            else if(id[p]==1) a[p]=max(1,a[p*2]);
            else a[p]=max(1,a[p*2+1]);
            p/=2;
        }
        cout<<a[1]<<endl;
    }
    return 0;
}