F. Maximum White Subtree

合集 - 思维题(99)

1.大雪球2024-01-162.E. Final Countdown2024-02-183.D. Divisible Pairs2024-02-194.F. Chat Screenshots2024-02-195.G. One-Dimensional Puzzle2024-02-206.P8775 [蓝桥杯 2022 省 A] 青蛙过河2024-02-217.Watering an Array2024-02-238.Dora and Search2024-02-249.P1137 旅行计划2024-02-2410.P3067 [USACO12OPEN] Balanced Cow Subsets G2024-02-2611.P4799 [CEOI2015 Day2] 世界冰球锦标赛2024-02-2612.A. Brick Wall2024-02-2813.B. Minimize Inversions2024-02-2814.C. XOR-distance2024-02-2815.A. Moving Chips2024-02-2816.B. Monsters Attack!2024-02-2817.C. Find B2024-02-2818.D. Slimes2024-02-2819.C. Turtle Fingers: Count the Values of k2024-02-2820.D. Turtle Tenacity: Continual Mods2024-02-2821.D. Vlad and Division2024-02-2822.C. LR-remainders2024-02-2823.P5322 [BJOI2019] 排兵布阵2024-03-0324.P2946 [USACO09MAR] Cow Frisbee Team S2024-03-0325.P1156 垃圾陷阱2024-03-0326.P1064 [NOIP2006 提高组] 金明的预算方案2024-03-0427.P3047 [USACO12FEB] Nearby Cows G2024-03-0428.P5020 [NOIP2018 提高组] 货币系统2024-03-0429.P3957 [NOIP2017 普及组] 跳房子2024-03-0630.P1550 [USACO08OCT] Watering Hole G2024-03-0631.P1525 [NOIP2010 提高组] 关押罪犯2024-03-0732.P1379 八数码难题2024-03-0733.P6121 [USACO16OPEN] Closing the Farm G2024-03-0834.P2341 [USACO03FALL / HAOI2006] 受欢迎的牛 G2024-03-0835.P9632 [ICPC2020 Nanjing R] K Co-prime Permutation2024-03-0836.P9825 [ICPC2020 Shanghai R] Fibonacci2024-03-0837.P2034 选择数字2024-03-0938.P7072 [CSP-J2020] 直播获奖2024-03-0939.P1714 切蛋糕2024-03-0940.P4147 玉蟾宫2024-03-1041.P2866 [USACO06NOV] Bad Hair Day S2024-03-1042.P2032 扫描2024-03-1043.P1884 [USACO12FEB] Overplanting S2024-03-1044.D. Lonely Mountain Dungeons2024-03-1245.F. Microcycle2024-03-1246.E. Increasing Subsequences__22024-03-1347.D. Valiant's New Map2024-03-1448.Array Repetition2024-03-1449.F. Rudolf and Imbalance2024-03-1450.D. Exam in MAC2024-03-1551.G. Rudolf and Subway2024-03-1552.C. Ehab and Path-etic MEXs2024-03-1753.A. String Transformation 12024-03-1754.D. Solve The Maze2024-03-1855.B. Array Fix2024-03-1856.B. Equal XOR2024-03-18

57.F. Maximum White Subtree2024-03-19

58.P1652 圆2024-03-2059.P2181 对角线2024-03-2060.P1325 雷达安装2024-03-2061.P3194 [HNOI2008] 水平可见直线2024-03-2062.B - Make Many Triangles2024-03-2363.[ARC172A] Chocolate2024-03-2564.C. Theofanis' Nightmare2024-03-2765.D. Birthday Gift2024-03-2766.C. Arrow Path2024-03-2767.P1265 公路修建2024-03-2868.P1284 三角形牧场2024-03-2869.Kirill and Mushrooms2024-03-2870.C. Sasha and the Casino2024-03-2871.P1470 [USACO2.3] 最长前缀 Longest Prefix2024-04-0172.P5536 【XR-3】核心城市2024-04-0273.P5836 [USACO19DEC] Milk Visits S2024-04-0274.P4391 [BOI2009] Radio Transmission 无线传输2024-04-0475.P3435 [POI2006] OKR-Periods of Words2024-04-0476.P5960 【模板】差分约束2024-04-0477.P3052 [USACO12MAR] Cows in a Skyscraper G2024-04-0578.E. Long Inversions2024-04-0979.C. Deep Down Below2024-04-0980.C. Rings2024-04-0981.F. Unfair Game2024-04-1082.G. GCD on a grid2024-04-1083.C. Inversion Graph2024-04-1084.Increase Subarray Sums2024-04-1085.D. Cross Coloring2024-04-1286.M. Triangle Construction2024-04-1287.D2. Set To Max (Hard Version)2024-04-1288.C. Array Game2024-04-1289.F. Alex's whims2024-04-1290.Queue Sort2024-04-1291.P1908 逆序对2024-04-1392.P5094 [USACO04OPEN] MooFest G 加强版2024-04-1593.P5149 会议座位2024-04-1594.P10337 [UESTCPC 2024] 操作序列2024-04-1595.P10330 [UESTCPC 2024] 黑白珠串2024-04-1596.P10252 线性变换2024-04-1697.P3901 数列找不同2024-04-1698.P10288 [GESP样题 八级] 区间2024-04-1699.P5677 [GZOI2017] 配对统计2024-04-17

收起

原题链接

题解

做两遍dfs，就能从树中求得图的效果了

第一遍，令 $f[i]$ 为以 $i$ 为根的子树（包括自身）的 $maxans$
则 $f[i]$ 的初值为

$\{\begin{array}{l}1,a[i]=1\\ -1,a[i]=0\end{array}$

$f[i]$ 的状态转移方程为 $f[i]+=f[u],f[u]>0,u\in So{n}_i$

第二遍怎么求呢？
已知根节点的 $f[i]$ 就已经是正确答案了，所以它所在的最小联通图上的节点的最小连通图就是根节点的最小连通图
$inotherwords$
如果当前子树大于零，那么它的父节点一定会用它，所以它的最小连通图相当于它父节点的最小连通图

code

// LUOGU_RID: 151626995
#include<bits/stdc++.h>
using namespace std;
map<int, vector<int> > G;
map<int,int> a;
map<int,int> f;

void ss_1(int now,int fa)
{
    f[now]=a[now];
    for(auto next:G[now])
    {
        if(next==fa) continue;

        ss_1(next,now);
        if(f[next]>0) f[now]+=f[next];
    }
}

void ss_2(int now,int fa)
{
    if(now!=fa)
    {
        if(f[now]>0) f[now]=max(f[now],f[fa]);
        else f[now]=max(f[now]+f[fa],f[now]);
    }
    for(auto next:G[now])
    {
        if(next!=fa) ss_2(next,now);
    }
}

int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        if(!a[i])a[i]=-1;
    }

    for(int i=1;i<n;i++)
    {
        int x,y;
        cin>>x>>y;
        G[x].emplace_back(y);
        G[y].emplace_back(x);
    }

    ss_1(1,1);

    ss_2(1,1);


    for(int i=1;i<=n;i++) cout<<f[i]<<" ";

    return 0;
}