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「ZOJ3626」Treasure Hunt I 树上背包

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字数统计: 536字 | 阅读时长 ≈ 3分钟

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题意

给一棵树每条边上有花费值$cost_i$,每个点上有权值$w_i$,求以$k$为起点,花费小于$m$,并最终回到起点$k$,能得到的最大权值

思路

树上的背包dp

令$f[i][j]$表示从$i$为起点,花费$j$所可以得到的最大权值.有两种决策对于$i$的孩子$v$,要么不走,要么花费$cost_{i \space to \space v}$

所以有$f[i][j]=max(f[i][j],f[v][k]+f[i][j-cost-k])$

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 = 110;
struct Edges{
    int to,nxt,cost;
}E[MAXN<<1];
int n,m,k,head[MAXN],w[MAXN],vis[MAXN],cnt,ans;
int f[MAXN][2010];
inline void addedge(int u,int v,int w){
    E[++cnt].nxt = head[u]; E[cnt].to = v; E[cnt].cost = w; head[u] = cnt;
}
void dfs(int u){
    vis[u]=1;
    f[u][0] = w[u];
    for (int i = head[u]; i; i = E[i].nxt){
        int v = E[i].to,cost = E[i].cost;
        if (!vis[v]) {
            dfs(v);
            per(j,m,0) rep(k,0,j-cost)
            f[u][j]=max(f[v][k]+f[u][j-k-cost],f[u][j]);
        }
    }
}
///------------------head------------------
signed main(signed argc, char *argv[]){
    while(scanf("%lld",&n) == 1){
        cnt=0;ans=-1;MM(vis,0);MM(head,0);MM(f,0);
        rep(i,1,n) w[i]=read();
        rep(i,1,n-1) {int u=read(),v=read(),w=read(); addedge(u,v,w); addedge(v,u,w);}
        k=read(),m=read()>>1;
        dfs(k);
        rep(i,0,m) ans=max(ans,f[k][i]);
        printf("%lld\n",ans);
    }
	return 0;
}