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「CF743D」Chloe and pleasant prizes

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字数统计: 465字 | 阅读时长 ≈ 3分钟

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题意

求最大的两子树权值之和,要求子树没有交集

思路

真·斯波题,直接统计最大子树和次大子树即可.

特判一条链的情况 是不可能有答案的.

恶心的inf弄我半天

Code

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//my vegetable has exploded. :(
#include<bits/stdc++.h>
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MM(x,y) memset(x,y,sizeof(x))
#define MCPY(a,b) memcpy(a,b,sizeof(b))
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,b,a) for(int i=b;i>=a;i--)
#define fi first
#define se second
#define int long long
using namespace std;

inline int quickpow(int m,int n,int p){int b=1;while(n){if(n&1)b=b*m%p;n=n>>1;m=m*m%p;}return b;}
inline int getinv(int x,int p){return quickpow(x,p-2,p);}
inline int read(void){
	int x=0,f=1;char ch=getchar();
	while(!isdigit(ch)){f=ch=='-'?-1:1;ch=getchar();}
	while(isdigit(ch)){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
	return x * f;
}
const int MAXN = 2e5+100;
const int inf = 0x3f3f3f3f3f3f3f3fLL;
int n,w[MAXN],fa[MAXN],dp[MAXN],head[MAXN],sz[MAXN],cnt,flag=0,ans=-inf;
struct Edges{
    int to,nxt;
}E[MAXN<<1];
inline void addedge(int u,int v){
    E[++cnt].nxt = head[u]; E[cnt].to = v; head[u] = cnt;
}
inline void dfs(int u,int f){
    int cur = -inf;
    fa[u] = f;
    dp[u] = -inf;
    for (int i = head[u]; i; i = E[i].nxt){
        int v = E[i].to;
        if (v == fa[u]) continue;
        ++sz[u];
        dfs(v,u);
        w[u] += w[v];
        dp[u] = max(dp[u],dp[v]);
        if (sz[u] > 1) flag = 1;
        ans = max(ans,cur + dp[v]);
        cur = max(cur,dp[v]);
    }
    dp[u] = max(dp[u],w[u]);
}
///------------------head------------------
signed main(signed argc, char *argv[]){
    n=read();
    rep(i,1,n) w[i]=read();
    rep(i,1,n-1) {int u=read(),v=read(); addedge(u,v); addedge(v,u);}
    dfs(1,-1);
    if (!flag) return puts("Impossible"),0;
    else return printf("%lld\n",ans),0;
}