hdu4276 依赖背包

Posted 2021-02-07 zsben991126

网上题解都是用spfa求1-n路径的，但其实dfs一次就可以了。。

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <iomanip>
 
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define maxn 1000005
#define MOD 1000000007
#define mem(a , b) memset(a , b , sizeof(a))
#define LL long long
#define INF 100000000
 
int n , t;
struct edge
{
    int u , v , w;
    int next;
}E[105*2];
int head[105];
int dp[105][505];
int val[105];
int cost[105] , per[105];
int id,sum ,flag ;
 
void add(int u , int v , int w)
{
    E[id].u = u;
    E[id].v = v;
    E[id].w = w;
    E[id].next = head[u];
    head[u] = id++;
}
/*
void spfa()
{
    mem(per , -1);
    for(int i = 2 ; i <= n ; i ++) cost[i] = INF;
    cost[1] = 0;
    queue<int>q;
    q.push(1);
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        for(int i = head[u] ; i >= 0 ; i = E[i].next)
        {
            if(cost[E[i].v] > cost[u] + E[i].w)
            {
                cost[E[i].v] = cost[u] + E[i].w;
                per[E[i].v] = i;
                q.push(E[i].v);
            }
        }
    }
    for(int i = 1 ; i <= n ; i ++) cout << per[i] << endl;
    for(int i = per[n] ; i >= 0 ; i = per[i])
    {
        cout << E[i].u << " " << E[i].v << " " << E[i].w << endl;
        E[i].w = 0;
    }
}*/
/*
bool judge(int u,int pre)///找出1~n的路径
{
    if(u == n)return true;
    for(int i = head[u]; i != -1 ; i = E[i].next)
    {
        int v = E[i].v;
        if(v == pre)continue;
        if(judge(v,u))
        {
            sum += E[i].w;
            E[i].w=0;
            return true;
        }
    }
    return false;///这句话必须有，因为这一句我没写WA到死.....
}
*/
 
int judge(int u , int per)
{
    int flag=0;if(u==n)return 1;
    for(int i = head[u] ; i >= 0 ; i = E[i].next )
    {
        if(E[i].v == per) continue;
        if(judge(E[i].v,u)) {sum += E[i].w ; E[i].w = 0 ; flag = 1 ;}
        else E[i].w*=2;
    }
    return flag;
}
 
void solve(int u , int per)
{
    for(int i = head[u] ; i >= 0 ; i = E[i].next)
    {
        if(E[i].v == per) continue;
        solve(E[i].v , u);
        for(int j = t ; j >= E[i].w ; j --)
        {
            int up = j - E[i].w;
            for(int k = 0 ; k <= up ; k ++)
            {
               // if(dp[u][up-k] != -1 && dp[E[i].v][k] != -1)
                    dp[u][j] = max(dp[u][j] , dp[u][up-k] + dp[E[i].v][k]);
            }
        }
    }
}
 
int main()
{
    while(scanf("%d %d" , &n , &t) != EOF)
    {
        mem(dp , 0);id = 0;
        mem(head , -1);
        int u , v , w;
        for(int i = 1 ; i < n ; i ++)
        {
            scanf("%d %d %d" , &u , &v , &w);
            //if(u > v) swap(u , v);
            add(u , v  , w);
            add(v , u , w);
        }
        for(int i = 1 ; i <= n ; i ++)
        {
            scanf("%d" , &val[i]);
            for(int j = 0 ; j <= t ; j ++) dp[i][j] = val[i];
        }
        flag = sum = 0;
        judge(1 , -1);
        //spfa();
        if(sum > t)
        {
            printf("Human beings die in pursuit of wealth, and birds die in pursuit of food!
");
            continue;
        }
        //cout << sum << endl;
        t -= sum;
        solve(1 , -1);
        printf("%d
" ,  dp[1][t]);
    }
}