ZOJ 4772 Treasure Hunt I 树形DP（背包） &&　hdu The Ghost Blows Light 树形DP（背包）

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4772

题意：

给定n个点n-1条边，每个点对应一个财富值，走每条路径都对应着一个所需要的时间，问在m天内从k出发然后回到k，所能取得到的最大财富值 v1 V2 ....V1.

思路：

由于题目给定的是一棵生成树，所以从k出发后必须按原路返回才可满足条件，也就是每条边走两次。

dp[k][m]表示从k出发的又回到k的路径所取得的最大值。就是相当于往容量为m的包里放物品一样。每个点对应一个包。

View Code

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>

#define CL(a,num) memset((a),(num),sizeof(a))
#define iabs(x)  ((x) > 0 ? (x) : -(x))
#define Min(a,b) (a) > (b)? (b):(a)
#define Max(a,b) (a) > (b)? (a):(b)

#define ll __int64
#define inf 0x7f7f7f7f
#define MOD 100000007
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define test puts("<------------------->")
#define maxn 100007
#define M 507
#define N 107
using namespace std;
//freopen("din.txt","r",stdin);

struct node{
int v,w;
int next;
}g
;
int head
,ct;
int dis
,dp
[M],val
;
int pre
,path
;
bool vt
;
int n,T;

void add(int u,int v,int w){
g[ct].v = v;
g[ct].w = w;
g[ct].next = head[u];
head[u] = ct++;
}
void spfa(){
queue<int>q;
int i;
CL(vt,false); CL(pre,-1);
for (i = 1; i <= n; ++i) dis[i] = inf;
dis[1] = 0;   vt[1] = true;
q.push(1);
while (!q.empty()){
int u = q.front(); q.pop();
for (i = head[u]; i != -1; i = g[i].next){
int v = g[i].v;
int w = g[i].w;
if (dis[v] > dis[u] + w){
pre[v] = u;//记录父节点
path[v] = i;//记录路径编号
dis[v] = dis[u] + w;
if (!vt[v]){
vt[v] = true;
q.push(v);
}
}
}
vt[u] = false;
}
for (i = n; i != 1; i = pre[i]){//置0
g[path[i]].w = 0;
}
}
void dfs(int u){
vt[u] = true;
for (int i = head[u]; i != -1; i = g[i].next){
int v = g[i].v;
int w = g[i].w*2;
if (vt[v]) continue;
dfs(v);
for (int k = T; k >= w; --k){
for (int j = 0; j <= k - w; ++j)
dp[u][k] = max(dp[u][k],dp[u][k - w - j] + dp[v][j]);//  这里当w为0时保证能够选入
}
}
for (int i = 0; i <= T; ++i) dp[u][i] += val[u];
}
int main(){
int i;
int u,v,w;
while (~scanf("%d%d",&n,&T)){
ct = 0;
CL(head,-1);
for (i = 2; i <= n; ++i){
scanf("%d%d%d",&u,&v,&w);
add(u,v,w);
add(v,u,w);
}
for (i = 1; i <= n; ++i) scanf("%d",&val[i]);
spfa();
//spfa求最短路后不满足条件
if (dis
> T){
printf("Human beings die in pursuit of wealth, and birds die in pursuit of food!\n");
continue;
}
T -= dis
;
CL(vt,false); CL(dp,0);
dfs(1);
printf("%d\n",dp[1][T]);
}
return 0;
}