HDU 5834 Magic boy Bi Luo with his excited tree 树形dp

题目链接：

http://acm.split.hdu.edu.cn/showproblem.php?pid=5834

Magic boy Bi Luo with his excited tree

Time Limit: 8000/4000 MS (Java/Others)
Memory Limit: 131072/131072 K (Java/Others) #### 问题描述 > Bi Luo is a magic boy, he also has a migic tree, the tree has N nodes , in each node , there is a treasure, it's value is V[i], and for each edge, there is a cost C[i], which means every time you pass the edge i , you need to pay C[i]. > > You may attention that every V[i] can be taken only once, but for some C[i] , you may cost severial times. > > Now, Bi Luo define ans[i] as the most value can Bi Luo gets if Bi Luo starts at node i. > > Bi Luo is also an excited boy, now he wants to know every ans[i], can you help him? #### 输入 > First line is a positive integer T(T≤104) , represents there are T test cases. > > Four each test： > > The first line contain an integer N(N≤105). > > The next line contains N integers V[i], which means the treasure’s value of node i(1≤V[i]≤104). > > For the next N−1 lines, each contains three integers u,v,c , which means node u and node v are connected by an edge, it's cost is c(1≤c≤104). > > You can assume that the sum of N will not exceed 106. #### 输出 > For the i-th test case , first output Case #i: in a single line , then output N lines , for the i-th line , output ans[i] in a single line. #### 样例 > **sample input** > 1 > 5 > 4 1 7 7 7 > 1 2 6 > 1 3 1 > 2 4 8 > 3 5 2 > > **sample output** > Case #1: > 15 > 10 > 14 > 9 > 15

题意

给你一颗树，每个点有价值，每个边有花费，价值只能取一次，花费每经过一次就要付一次，问从编号为x(x=1,...,n)，能够获得的最大价值是多少。

题解

dp[u][0]:从u往下走能回来的最大价值
dp[u][1]:从u往下走不回来的最大价值（也有可能根本不往下走哦），d[u]:从哪个位置出去不回来的
dp[u][2]:从u往下走不回来的次大价值
这些在第一次dfs的时候就能求出来。

令up1表示从父亲出去不回来的最大价值,up2表示从父亲出去要回来的最大价值
那么ans[i]=max(up1+dp[u][0],up2+dp[u][1])
所以在dfs2中维护下up1,up2，就能往下推出所有的ans了。

代码

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=1e5+10;

int dp[maxn][3],id[maxn];
int val[maxn],ans[maxn];

VPII G[maxn];

//dp[u][0]:从u往下走能回来的最大价值
//dp[u][1]:从u往下走不回来的最大价值（也有可能根本不往下走哦），d[u]:从哪个位置出去不回来的
//dp[u][2]:从u往下走不回来的次大价值
void dfs(int u,int fa) {
    dp[u][0]=dp[u][1]=dp[u][2]=val[u];
    id[u]=-1;
    rep(i,0,G[u].sz()) {
        int v=G[u][i].X,w=G[u][i].Y;
        if(v==fa) continue;
        dfs(v,u);
        int tmp=dp[v][0]-2*w;
        if(tmp>0) dp[u][0]+=tmp;
    }
    rep(i,0,G[u].sz()) {
        int v=G[u][i].X,w=G[u][i].Y;
        if(v==fa) continue;
        int tmp=dp[u][0]-max(0,(dp[v][0]-2*w))+max(0,dp[v][1]-w);
        if(tmp>dp[u][1]) {
            dp[u][2]=dp[u][1];
            dp[u][1]=tmp;
            id[u]=v;
        } else if(tmp>dp[u][2]) {
            dp[u][2]=tmp;
        }
    }
}

//dfs2磨了很久，还是参考别人写出来的：http://www.cnblogs.com/hchlqlz-oj-mrj/p/5774196.html
//当时一直想维护出全局的值，然后就晕了。。xrz
//up1表示从父亲出去不回来的最大价值,up2表示从父亲出去要回来的最大价值
void dfs2(int u,int fa,int up1,int up2) {
    ans[u]=max(up1+dp[u][0],up2+dp[u][1]);
    rep(i,0,G[u].sz()){
        int v=G[u][i].X,w=G[u][i].Y;
        if(v==fa) continue;
        int dwn2=up2+dp[u][0]-max(0,dp[v][0]-2*w);
        dwn2=max(0,dwn2-2*w);
        int dwn1=up1+dp[u][0]-max(0,dp[v][0]-2*w);
        dwn1=max(0,dwn1-w);
        if(id[u]!=v){
            int tmp=up2+dp[u][1]-max(0,dp[v][0]-2*w);
            tmp=max(0,tmp-w);
            dwn1=max(dwn1,tmp);
        }else{
            int tmp=up2+dp[u][2]-max(0,dp[v][0]-2*w);
            tmp=max(0,tmp-w);
            dwn1=max(dwn1,tmp);
        }
        dfs2(v,u,dwn1,dwn2);
    }
}

void init(){
    rep(i,0,maxn) G[i].clear();
}

int main() {
    int tc,kase=0;
    int n;
    scanf("%d",&tc);
    while(tc--) {
        scf("%d",&n);
        init();
        for(int i=1; i<=n; i++) scf("%d",&val[i]);
        rep(i,1,n) {
            int u,v,w;
            scf("%d%d%d",&u,&v,&w);
            G[u].pb(mkp(v,w));
            G[v].pb(mkp(u,w));
        }

        dfs(1,-1);
        dfs2(1,-1,0,0);

        prf("Case #%d:\n",++kase);
        for(int i=1; i<=n; i++) {
            prf("%d\n",ans[i]);
        }
    }
    return 0;
}

//end-----------------------------------------------------------------------