【树形DP---未解决】hdu 3660

来源：互联网发布：网页美工属于web什么编辑：程序博客网时间：2024/06/05 08:47

题意：给出n个点，n-1条边，问从节点0到最远的子叶的距离满足L<=dis<=R，且Bob总是选最大的，Alice总是选最小的，问最后能否满足要求。

方法：树形DP，dis[u]数组表示0节点到当前节点u的距离，dp[u]数组表示最远的子叶节点到u节点的满足要求的距离，最后dp[0]就是答案。这dp妙就妙在利用了的dfs回溯，从最远子节点到0节点进行dp

贴一个TLE代码，有空a了他！

#include <list>#include <map>#include <set>#include <queue>#include <string>#include <deque>#include <stack>#include <algorithm>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <limits.h>#include <time.h>#include <string.h>using namespace std;#define LL long long#define PI acos(-1.0)#define MAX INT_MAX#define MIN INT_MIN#define eps 1e-8#define FRE freopen("a.txt","r",stdin)#define MOD 1000000007#define N 500010int max(int a,int b){return a>b?a:b;}int min(int a,int b){return a>b?b:a;}struct edge{    int to;    int val;};vector<edge> v[N];int dis[N],dp[N];int l,r;void dfs(int u,int who){    int i,j;    if(v[u].size()==0){dp[u]=0;return ;}    dp[u]= who?0:MAX;    if(dis[u]>r)return ;    for(i=0;i<v[u].size();i++){        int to=v[u][i].to;        int w=v[u][i].val;        dis[to]=dis[u]+w;        dfs(to,!who);        if(dis[u]+w+dp[to]<=r && dis[u]+w+dp[to]>=l){            if(who)            dp[u]=max(dp[u],dp[to]+w);            else            dp[u]=min(dp[u],dp[to]+w);        }    }}int main(){    int n;    while(~scanf("%d%d%d",&n,&l,&r)){        int i,j,k;        for(i=0;i<=n;i++)v[i].clear();        for(i=1;i<n;i++){            int a,b,c;            scanf("%d%d%d",&a,&b,&c);            edge tmp;            tmp.to=b;            tmp.val=c;            v[a].push_back(tmp);        }        dis[0]=0;        dfs(0,1);        if(dp[0]<=r && dp[0]>=l)printf("%d\n",dp[0]);        else        printf("Oh, my god!\n");    }    return 0;}