hdu4616（树形dp）

最新推荐文章于 2024-05-22 17:32:19 发布

转载最新推荐文章于 2024-05-22 17:32:19 发布 · 147 阅读

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/martinue/p/5490423.html

文章标签：

#java

本文解析了一道关于游戏挑战的树形DP题目，玩家在一个类似树状结构的空间内收集礼物，目标是在有限的陷阱触发次数下获得最大礼物价值。

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Game

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1899 Accepted Submission(s): 604

Problem Description

　　Nowadays, there are more and more challenge game on TV such as 'Girls, Rush Ahead'. Now, you participate int a game like this. There are N rooms. The connection of rooms is like a tree. In other words, you can go to any other room by one and only one way. There is a gift prepared for you in Every room, and if you go the room, you can get this gift. However, there is also a trap in some rooms. After you get the gift, you may be trapped. After you go out a room, you can not go back to it any more. You can choose to start at any room ,and when you have no room to go or have been trapped for C times, game overs. Now you would like to know what is the maximum total value of gifts you can get.

Input

　　The first line contains an integer T, indicating the number of testcases.
　　For each testcase, the first line contains one integer N(2 <= N <= 50000), the number rooms, and another integer C(1 <= C <= 3), the number of chances to be trapped. Each of the next N lines contains two integers, which are the value of gift in the room and whether have trap in this rooom. Rooms are numbered from 0 to N-1. Each of the next N-1 lines contains two integer A and B(0 <= A,B <= N-1), representing that room A and room B is connected.
　　All gifts' value are bigger than 0.

Output

　　For each testcase, output the maximum total value of gifts you can get.

Sample Input

Sample Output

146
158

题意：

给一棵树，每个结点中有礼物，每个礼物有一个权值，某些结点中会有陷阱，你可以从任何一点出发，每个结点最多只能经过一次，最多掉进陷阱C次，求出可获得的礼物的最大值。

树形dp，也可以直接dfs，不过这题数据太bug。。

暴力：

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
using namespace std;
typedef long long ll;
const int N = 50010,INF=999999999;
vector<int> tu[N];
int val[N];
int  flag[N];
bool vis[N];
int n,c;
int dfs(int x,int pre,int cc)
{
    if(cc-flag[x]<=0)return val[x];
    else cc-=flag[x];
    int l=tu[x].size(),ans=val[x],max0=0;
    for(int i=0; i<l; i++)
    {
        int t=tu[x][i];
        if(t==pre)continue;
        max0=max(max0,dfs(t,x,cc));
    }
    return ans+max0;
}

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        scanf("%d%d",&n,&c);
        for(int i=0; i<n; i++)
            tu[i].clear();
        memset(flag,0,sizeof(flag));
        for(int i=0; i<n; i++)
            scanf("%d%d",&val[i],&flag[i]);
        for(int i=0; i<n-1; i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            tu[a].push_back(b);
            tu[b].push_back(a);
        }
        int ans=-1;
        for(int i=0; i<n; i++)
            if(tu[i].size()==1)
                ans=max(ans,dfs(i,-1,c));
        printf("%d\n",ans);
    }
}

用dp[u][j]表示以u为根的子树上，从某点走到u且走过j个陷阱能得到的最大值;

考虑如果最优解走过了k个陷阱，如果k!=C，那么其起点和终点都可以为没有陷阱的点，如果k=C，那么起点或者终点至少有一点为有陷阱的点，所以dp数组还需要增加一维：dp[u][i][flag]，若flag=1，表示以u为根的子树上，从一个有陷阱的点走到u走过j个陷阱能得到的最大值，若flag=0，表示起点没有陷阱。

状态转移：

1) 当u点本身就有陷阱时，dp[u][j+1][flag]=max{dp[v][j][flag]+val[u]}，0<=j<C，v是u的儿子;

2) 当u点没有陷阱时，dp[u][j][flag]=max{dp[v][j][flag]+val[u]}，0<=j<C，特殊的，当j=C时，因为路上走过了C个陷阱，所以起点和终点不能同时为没有陷阱的点，所以只有flag=1，能用上式转移，也就是说dp[u][C][0]是不可能的情况。

这两种情况可以合成一种写。

更新ans：最优解可以看成两条链拼在一起，枚举两条链上陷阱的个数然后再拼起来。

树形dp：

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
using namespace std;
typedef long long ll;
#define maxn 50005
const long long INF=999999999;
int n,c;
vector<int> tu[maxn];
int val[maxn],flag[maxn];
int ans,dp[maxn][4][2];
void dfs(int u,int p)
{
    int s=tu[u].size();
    dp[u][flag[u]][flag[u]]=val[u];
    ans=max(ans,dp[u][flag[u]][flag[u]]);
    for (int i=0; i<s; i++)
    {
        int v=tu[u][i];
        if (v==p) continue;
        dfs(v,u);
        for (int j=0; j<=c; j++)
            for (int k=0; k+j<=c; k++)
            {
                if (k+j<=c) ans=max(ans,dp[u][j][1]+dp[v][k][1]);
                if (k+j<c)  ans=max(ans,dp[u][j][0]+dp[v][k][0]);
                if (j!=c)   ans=max(ans,dp[u][j][0]+dp[v][k][1]);
                if (k!=c)   ans=max(ans,dp[u][j][1]+dp[v][k][0]);
            }
        for (int j=0; j<c; j++)
        {
            dp[u][j+flag[u]][0]=max(dp[u][j+flag[u]][0],dp[v][j][0]+val[u]);
            dp[u][j+flag[u]][1]=max(dp[u][j+flag[u]][1],dp[v][j][1]+val[u]);
        }
        if (!flag[u]) dp[u][c][1]=max(dp[u][c][1],dp[v][c][1]+val[u]);
    }
}
int main()
{
    int t,a,b;
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d%d",&n,&c);
        for (int i=0; i<n; i++)
            tu[i].clear();
        for (int i=0; i<n; i++)
            scanf("%d%d",&val[i],&flag[i]);
        for (int i=1; i<n; i++)
        {
            scanf("%d%d",&a,&b);
            tu[a].push_back(b);
            tu[b].push_back(a);
        }
        ans=0;
        for (int i=0; i<n; i++)
            for (int j=0; j<4; j++)
                for (int k=0; k<2; k++)
                    dp[i][j][k]=-INF;
        dfs(0,-1);
        printf("%d\n",ans);
    }
    return 0;
}

转载于:https://www.cnblogs.com/martinue/p/5490423.html