HDOJ question 4276 The Ghost Blows Light (SPFA + tree DP)

HDOJ question 4276 The Ghost Blows Light (SPFA + tree DP)
The Ghost Blows LightTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 2549 Accepted Submission (s): 795


Problem Description
My name is Hu Bayi, robing an existing ent tomb in Tibet. the tomb consists of N rooms (numbered from 1 to N) which are connected by some roads (pass each road shocould 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 shocould 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,! = 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 Input

5 101 2 2 2 3 22 5 33 4 31 2 3 4 5

Sample Output

11

Source 2012 ACM/ICPC Asia Regional Changchun Online
Recommend liuyiding | We have carefully selected several similar problems for you: 4272 4277 4274 4269 theme: a tree, from to, arrives within t time, the biggest value obtained: first use spfa to find the shortest path. If it is greater than t, it cannot be reached. t minus the shortest path, the edge weight on the shortest path is changed to 0, if the path is not the shortest path, the back and forth are twice the edge weight, that is, the consumption of the edge weight ac code is twice.

#include
 
  #include
  
   #include
   
    #include
    
     #include
     
      #define max(a,b) (a>b?a:b)#define INF 0xfffffffusing namespace std;int n,t;int a[110],head[110],vis[110],cnt,pre[110],dis[110],d[220],dp[110][10100];struct s{int u,v,w,next;}edge[220];void add(int u,int v,int w){edge[cnt].u=u;edge[cnt].v=v;edge[cnt].w=w;edge[cnt].next=head[u];head[u]=cnt++;}void spfa(int u){int i;for(i=1;i<=n;i++){dis[i]=INF;pre[i]=i;d[i]=0;}memset(vis,0,sizeof(vis));vis[u]=1;dis[u]=0;queue
      
       q;q.push(u);while(!q.empty()){int u=q.front();q.pop();vis[u]=0;for(i=head[u];i!=-1;i=edge[i].next){int v=edge[i].v;int w=edge[i].w;if(dis[v]>dis[u]+w){dis[v]=dis[u]+w;pre[v]=u;d[v]=i;if(!vis[v]){vis[v]=1;q.push(v);}}}}for(i=n;i!=1;i=pre[i]){edge[d[i]].w=0;edge[d[i]^1].w=0;}}void dfs(int u,int pre){int i,j,k;for(i=0;i<=t;i++){dp[u][i]=a[u];}for(i=head[u];i!=-1;i=edge[i].next){int v=edge[i].v;if(v==pre)continue;dfs(v,u);int cost=edge[i].w*2;for(j=t;j>=cost;j--){for(k=0;k+cost<=j;k++){dp[u][j]=max(dp[u][j],dp[v][k]+dp[u][j-k-cost]);}}}}int main(){while(scanf(%d%d,&n,&t)!=EOF){int i;memset(head,-1,sizeof(head));memset(dp,0,sizeof(dp));cnt=0;for(i=0;i
       
        t){printf(Human beings die in pursuit of wealth, and birds die in pursuit of food!);continue;}t-=dis[n];dfs(1,-1);printf(%d,dp[1][t]);}}