題目鏈接

POJ3585

題解

-二次掃描與換根法-
對於這樣一個無根樹的樹形dp
我們先任選一根進行一次樹形dp
然後再掃一遍通過計算得出每個點為根時的答案

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define LL long long int
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
 

#define cls(s) memset(s,0,sizeof(s))
using namespace std;
const int maxn = 200005,maxm = 400005,INF = 1000000000;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == ‘-‘) flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3
 
) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
int h[maxn],ne = 2,de[maxn];
struct EDGE{int to,nxt,w;}ed[maxm];
inline void build(int u,int v,int w){
    ed[ne] = (EDGE){v,h[u],w}; h[u] = ne++;
    ed[ne] = (EDGE){u,h[v],w}; h[v] = ne++;
    de[u]++; de[v]++;
}
int n,fa[maxn];
int
 
 d[maxn],f[maxn],g[maxn],ans;
void dfs1(int u){
    f[u] = 0;
    if (de[u] == 1 && u != 1) return;
    Redge(u) if ((to = ed[k].to) != fa[u]){
        fa[to] = u; d[to] = ed[k].w; dfs1(to);
        if (de[to] == 1) f[u] += ed[k].w;
        else f[u] += min(ed[k].w,f[to]);
    }
}
void dfs2(int u){
    g[u] = f[u];
    if (fa[u]){
        if (de[fa[u]] == 1) g[u] += d[u];
        else g[u] += min(d[u],g[fa[u]] - min(d[u],f[u]));
    }
    ans = max(ans,g[u]);
    Redge(u) if ((to = ed[k].to) != fa[u]){
        dfs2(to);
    }
}
int main(){
    int T = read();
    while (T--){
        ne = 2; cls(h); cls(de); ans = 0;
        n = read(); int a,b,w;
        for (int i = 1; i < n; i++){
            a = read(); b = read(); w = read();
            build(a,b,w);
        }
        dfs1(1);
        //REP(i,n) printf("%lld ",f[i]); puts("");
        dfs2(1);
        //REP(i,n) printf("%lld ",g[i]); puts("");
        printf("%d\n",ans);
    }
    return 0;
}

POJ3585 Accumulation Degree 【樹形dp】