HDOJ 4276 The Ghost Blows Light 树上分组背包

先判断从1号点能否顺利到n号点,并记录路径,和所需的时间.

然后从1号点开始跑树上的分组背包,需要注意的是:
1. 原路径上的点的权值提前加出来.
2. 路径上的两个点间的边长重置为0.

dp方程:
设v点为u的孩子,c为u到v的边长,dp[u][vl]表示在u点花了vl的时间,则v到u的转移为:

dp[u][vl]=max(dp[u][vl],dp[u][vl−j]+dp[v][j−c])

u可能有很多个孩子,依次按上面转移跑很多遍01背包即可.

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The Ghost Blows Light

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2788 Accepted Submission(s): 862

Problem Description

My name is Hu Bayi, robing an ancient tomb in Tibet. The tomb consists of N rooms (numbered from 1 to N) which are connected by some roads (pass each road should cost some time). There is exactly one route between any two rooms, and each room contains some treasures. Now I am located at the 1st room and the exit is located at the Nth room.
Suddenly, alert occurred! The tomb will topple down in T minutes, and I should reach exit room in T minutes. Human beings die in pursuit of wealth, and birds die in pursuit of food! Although it is life-threatening time, I also want to get treasure out as much as possible. Now I wonder the maximum number of treasures I can take out in T minutes.

Input

There are multiple test cases.
The first line contains two integer N and T. (1 <= n <= 100, 0 <= T <= 500)
Each of the next N - 1 lines contains three integers a, b, and t indicating there is a road between a and b which costs t minutes. (1<=a<=n, 1<=b<=n, a!=b, 0 <= t <= 100)
The last line contains N integers, which Ai indicating the number of treasure in the ith room. (0 <= Ai <= 100)

Output

For each test case, output an integer indicating the maximum number of treasures I can take out in T minutes; if I cannot get out of the tomb, please output “Human beings die in pursuit of wealth, and birds die in pursuit of food!”.

Sample Input5 101 2 2 2 3 22 5 33 4 31 2 3 4 5Sample Output11

Source
2012 ACM/ICPC Asia Regional Changchun Online

/* ***********************************************Author        :CKbossCreated Time  :2015年09月09日 星期三 13时46分59秒File Name     :HDOJ4276.cpp************************************************ */#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <cmath>#include <cstdlib>#include <vector>#include <queue>#include <set>#include <map>using namespace std;const int maxn=200;const int inf=0x3f3f3f3f;int n,T;struct Edge{    int to,next,cost;}edge[maxn*2];int Adj[maxn],Size;void init(){    memset(Adj,-1,sizeof(Adj)); Size=0;}void Add_Edge(int u,int v,int c){    edge[Size].to=v;    edge[Size].next=Adj[u];    edge[Size].cost=c;    Adj[u]=Size++;}int dist[maxn],pre[maxn];int valu[maxn];void dfs(int u,int fa){    pre[u]=fa;    for(int i=Adj[u];~i;i=edge[i].next)    {        int v=edge[i].to;        int c=edge[i].cost;        if(v==fa) continue;        dist[v]=dist[u]+c;        dfs(v,u);    }}bool onroad[maxn];int ans;int findRoad(){    memset(onroad,false,sizeof(onroad));    int u=n,ret=0;    while(u!=pre[u]) { ret+=valu[u]; valu[u]=0; onroad[u]=true; u=pre[u]; }    onroad[1]=true;    ret+=valu[1]; valu[1]=0;    return ret;}int dp[maxn][5*maxn];//// fenzhu beibaovoid DP(int u,int fa){    for(int i=Adj[u];~i;i=edge[i].next)    {        int v=edge[i].to;        int c=edge[i].cost;        if(v==fa) continue;        DP(v,u);        if(onroad[u]&&onroad[v]) c=0;        for(int vl=T;vl>=0;vl--) /// 容量        {            for(int j=vl;j>=0;j--) /// 花费时间            {                dp[u][vl]=max(dp[u][vl],dp[u][vl-j]+dp[v][j-c]);            }        }    }}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    while(scanf("%d%d",&n,&T)!=EOF)    {        init();        for(int i=0;i<n-1;i++)        {            int u,v,c;            scanf("%d%d%d",&u,&v,&c);            Add_Edge(u,v,c); Add_Edge(v,u,c);        }        for(int i=1;i<=n;i++)         {            scanf("%d",valu+i); pre[i]=i;            ///init dp        }        memset(dist,63,sizeof(dist)); dist[1]=0;        dfs(1,1);        if(dist[n]>T)        {            puts("Human beings die in pursuit of wealth, and birds die in pursuit of food!");            continue;        }        //// find road        ans=findRoad();        T-=dist[n]; T/=2;        memset(dp,0,sizeof(dp));        for(int i=1;i<=n;i++)        {            for(int j=0;j<=T;j++)             {                if(onroad[i]==false) dp[i][j]=valu[i];            }        }        DP(1,1);        printf("%d\n",dp[1][T]+ans);    }    return 0;}