HDU-4616 Game 树形DP

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4616

　　比较典型的树形DP题目，f[u][j][k]表示以点u为子树，经过 j 个陷阱的最大值，其中k=0表示从u点出发，k=1表示终点为点u。则转移方程为：f[u][j+is_rtap][k]=Max{ f[v][j][k] | v为u的儿子节点,0<=j<=m }。要分别对每颗子树求出最大值，枚举其中的两颗子树，一颗出，一颗进，更新最大值。要注意在更新k=0时的最大值是，j要从1开始遍历，因为如果当前u节点存在trap，而u中的其中一个子节点v没有trap，那么会使f[u][1][0]加上f[v][0][0]的值，而不满足题目中遇到满足的m后停止下来。

//STATUS:C++_AC_93MS_5540KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
#pragma comment(linker,"/STACK:102400000,102400000")
using namespace std;
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef __int64 LL;
typedef unsigned __int64 ULL;
//const
const int N=50010,M=2000010;
const int INF=0x3f3f3f3f;
const int MOD=100000,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

struct Edge{
    int u,v;
}e[N<<1];
int first[N],next[N<<1],tr[N];
int T,n,m,mt;
LL f[N][4][2],w[N],ans;

void adde(int a,int b)
{
    e[mt].u=a,e[mt].v=b;
    next[mt]=first[a],first[a]=mt++;
    e[mt].u=b,e[mt].v=a;
    next[mt]=first[b],first[b]=mt++;
}

void dfs(int u,int fa)
{
    int i,j,k,v;
    for(i=first[u];~i;i=next[i]){
        v=e[i].v;
        if(v==fa)continue;
        dfs(v,u);
        for(j=0;j<=m;j++){
            for(k=0;k+j<=m;k++){
                if(j<m)ans=Max(ans,f[u][j][1]+f[v][k][0]);
                if(k<m)ans=Max(ans,f[u][j][0]+f[v][k][1]);
                if(j+k<m)ans=Max(ans,f[u][j][1]+f[v][k][1]);
            }
        }
        for(j=0;j+tr[u]<=m;j++){
            if(j && tr[u]<m)f[u][j+tr[u]][0]=Max(f[u][j+tr[u]][0],f[v][j][0]+w[u]);
            if(j<m)f[u][j+tr[u]][1]=Max(f[u][j+tr[u]][1],f[v][j][1]+w[u]);
        }
    }
}

int main()
{
 //   freopen("in.txt","r",stdin);
    int i,j,a,b;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        mem(f,0);
        for(i=0;i<n;i++){
            scanf("%I64d%d",&w[i],&tr[i]);
            f[i][tr[i]][0]=f[i][tr[i]][1]=w[i];
        }
        mt=0;mem(first,-1);
        for(i=1;i<n;i++){
            scanf("%d%d",&a,&b);
            adde(a,b);
        }

        ans=0;
        dfs(0,-1);

        printf("%I64d
",ans);
    }
    return 0;
}