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最新推荐文章于 2025-06-15 10:05:58 发布

原创最新推荐文章于 2025-06-15 10:05:58 发布 · 309 阅读

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本文探讨了一种解决特定图论问题的方法，该问题涉及寻找最优路径和子图，通过使用深度优先搜索和动态规划技术来优化解决方案。文章提供了一个完整的C++实现案例，包括全局变量误用和算法细节解释。

昨天吐槽还没A，今天就A了
有个变量开成了全局变量，应该携程局部变量

对于中间的solve我也不懂为什么是nlog2n，我不看题解也不会做

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int INF = 2000000000;
const int MAXN = 4e5+5;

int ans;
int co[MAXN];
int vis[MAXN];
struct Pode{
    int to,nx, di;
    Pode(int a=0,int b=0,int c=0):to(a), nx(b), di(c){}
}E[MAXN];
int head[MAXN], cot;
void add(int u, int v, int w) {
    E[cot] = Pode(v,head[u],w); head[u] = cot++;
}
int N,K,M;
/***************WeightRoot************/
int all, num, center;
int dp[MAXN], nodes[MAXN]; 
void findRoot(int x,int pre) {
    nodes[x] = 1; dp[x] = 0;
    for(int i = head[x]; ~i; i = E[i].nx) {
        int y = E[i].to; if(y == pre || vis[y]) continue;
        findRoot(y,x);
        nodes[x] += nodes[y];
        dp[x] = max(dp[x], nodes[y]);
    }
    dp[x] = max(dp[x], all-nodes[x]);
    if(dp[x] < num) {
        num = dp[x]; center = x;
    }
}
int getRoot(int root,int sn) {
    num = INF; all = sn; center = root;
    findRoot(root, -1);
    return center;  
}
/*************treecdq***************/
struct Node{
    int dep, v, di;
}so[MAXN]; 
int cmp(Node a,Node b) {
    return a.dep < b.dep;
}
int dep[MAXN]; // max dep (black)
int g[MAXN]; 
int mg[MAXN];
void getdep(int x,int pre) {
    dep[x] = co[x]; int res = 0;
    for(int i = head[x]; ~i; i = E[i].nx) {
        int y = E[i].to; if(y == pre || vis[y]) continue;
        getdep(y,x);
        res = max(res, dep[y]);
    }   
    dep[x] += res;
}
void getg(int x,int pre,int d,int c) {
    g[c] = max(g[c], d);
    for(int i = head[x]; ~i; i = E[i].nx) {
        int y = E[i].to; if(y == pre || vis[y]) continue;
        getg(y,x,d+E[i].di, c+co[y]);
    }
}
void work(int x) {
    vis[x] = 1; int tot = 0;
    for(int i = head[x]; ~i; i = E[i].nx) {
        int y = E[i].to; if(vis[y]) continue;
        work(getRoot(y,nodes[y]));
    }
    for(int i = head[x]; ~i; i = E[i].nx) {
        int y = E[i].to; if(vis[y]) continue;
        getdep(y, x);
        so[++tot].dep = dep[y]; so[tot].v = y; so[tot].di = E[i].di;
    }
    sort(so+1,so+tot+1,cmp);
//  printf("%d:",x); for(int i = 1; i <= tot; ++i) printf("%d ",so[i].dep); printf("\n");
    for(int i = 0; i <= so[tot].dep; ++i) mg[i] = -INF;
    for(int i = 1; i <= tot; ++i) {
        int t1 = so[i].dep; int t2 = so[i].v; int t3 = so[i].di;
        for(int j = 0; j <= t1; ++j) g[j] = -INF;
        getg(t2,x,t3,co[t2]);
        if(i != 1) {
            for(int j = 0; j <= K-co[x] && j <= t1; ++j) {
                int tt = min(so[i-1].dep, K-co[x]-j);
                if(mg[tt] == -INF) break;
                if(g[j] != -INF) ans = max(ans, mg[tt]+g[j]);
            }
        }
        for(int j = 0; j <= t1; ++j) {
            mg[j] = max(g[j], mg[j]);
            if(j) mg[j] = max(mg[j], mg[j-1]);
            if(j+co[x] <= K) ans=max(ans, mg[j]);   
        }
    }
    vis[x] = 0;
}

int main(){ 
    while(~scanf("%d %d %d",&N,&K,&M)) {
        ans = 0;
        memset(head,-1,sizeof(head)); cot = 0;
        memset(co,0,sizeof(co));
        memset(vis,0,sizeof(vis));
        for(int i = 1; i <= M; ++i) {
            int a; scanf("%d",&a);
            co[a] ++;
        }
        for(int i = 1; i < N; ++i) {
            int a,b,c; scanf("%d %d %d",&a,&b,&c);
            add(a,b,c); add(b,a,c);
        }
        work(getRoot(1,N));
        printf("%d\n",ans);
    }
    return 0;
}