hdu 4276 The Ghost Blows Light(DP-树形DP)

The Ghost Blows Light

[b]Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2240 Accepted Submission(s): 688

[/b]

[align=left]Problem Description[/align]

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.

[align=left]Input[/align]
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)

[align=left]Output[/align]
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!".

[align=left]Sample Input[/align]

5 10
1 2 2
2 3 2
2 5 3
3 4 3
1 2 3 4 5

[align=left]Sample Output[/align]

11

#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 110;
const int maxt = 510;
int dp[maxn][maxt] , A[maxn] , vis[maxn] , ans[maxt];
int way[maxn];
int N , T , cnt , sumT , route;

struct edge{
int u , v , t , next;
edge(int a = 0 , int b = 0, int c = 0 , int d = 0){
u = a , v = b , t = c , next = d;
}
}e[4*maxn];
int head[maxn];

void add(int u , int v , int t){
e[cnt] = edge(u , v , t , head[u]);
head[u] = cnt++;
}

void initial(){
for(int i = 0; i < maxn; i++){
head[i] = -1;
A[i] = 0;
vis[i] = 0;
for(int j = 0; j < maxt; j++) dp[i][j] = 0;
}
for(int i = 0; i < maxt; i++) ans[i] = 0;
cnt = 0;
sumT = 0;
route = 0;
}

void readcase(){
int u , v , t;
for(int i = 0; i < N-1; i++){
scanf("%d%d%d" , &u , &v , &t);
add(u , v , t);
add(v , u , t);
}
for(int i = 1; i <= N; i++){
scanf("%d" , &A[i]);
}
}

void DP(int k , int f){
dp[k][0] = A[k];
for(int i = head[k]; i != -1; i = e[i].next){
int son = e[i].v;
if(son != f){
DP(son , k);
if(vis[son]) continue;
for(int t = T; t >= 0; t--){
if(dp[k][t] == 0 && t != 0) continue;
for(int j = T; j >= 0; j--){
if(dp[son][j] != 0 && t+j+2*e[i].t <= T && dp[k][t+j+2*e[i].t] < dp[k][t]+dp[son][j]) dp[k][t+j+2*e[i].t] = dp[k][t]+dp[son][j];
}
}
}
}
}

bool dfs(int k , int index , int f){
way[index] = k;
route = index;
if(k == N) return true;
for(int i = head[k]; i != -1; i = e[i].next){
int son = e[i].v;
if(f != son){
sumT += e[i].t;
if(dfs(son , index+1 , k)) return true;
sumT -= e[i].t;
}
}
return false;
}

void computing(){
dfs(1 , 0 , 1);
for(int i = 0; i <= route; i++) vis[way[i]] = 1;
DP(1 , 1);
if(sumT > T){
printf("Human beings die in pursuit of wealth, and birds die in pursuit of food!\n");
return;
}
T -= sumT;
for(int i = 0; i <= route; i++){
for(int t = T; t >= 0; t--){
if(ans[t] == 0 && t != 0) continue;
for(int j = T; j >= 0; j--){
if(t+j <= T && ans[t+j] < ans[t]+dp[way[i]][j]) ans[t+j] = ans[t]+dp[way[i]][j];
}
}
}
int Max = 0;
for(int i = 0; i <= T; i++) if(Max < ans[i]) Max = ans[i];
printf("%d\n" , Max);
}

int main(){
while(~scanf("%d%d" , &N , &T)){
initial();
readcase();
computing();
}
return 0;
}