UVALive-5088 Alice and Bob & amp; #39; s Trip

A and B are going to travel. There are N cities. The roads in these cities are a tree. And the edges are directed. A and B are traveling from 0. A is too lazy to walk as little as possible. B is very energetic and wants to walk as much as possible, but no matter how you choose their total journey, it must be within the range of [L, R. So they choose the path that they choose. Start from B. Then you will be asked if you adopt the optimal strategy. The maximum number of steps B can perform.

Idea: Use dp [I] to indicate that the node I can obtain the maximum value. A tree-like DP is not difficult. Note that the length of the node from the beginning to the present is recorded, in addition, it is relatively simple to select B first and then select B in turn.

#include 
 
  #include 
  
   #include 
   
    #include #include 
    
     using namespace std;const int INF = 0x3f3f3f3f;const int MAXN = 500050;struct edge{    int v,w;    edge(){}    edge(int vv,int ww):v(vv),w(ww){}};vector
     
       arr[MAXN];int L,R,n;int dp[MAXN];void dfs(int x,int cnt,int pres){    dp[x] = 0;    if (arr[x].size() == 0)        return;    int temp = 0;    if (cnt & 1){        temp = INF;        for (int i = 0; i < arr[x].size(); i++){            int y = arr[x][i].v;            dfs(y,cnt+1,pres+arr[x][i].w);            if (pres + dp[y] + arr[x][i].w >= L && pres + dp[y] + arr[x][i].w <= R)                temp = min(temp,dp[y]+arr[x][i].w);        }        dp[x] = temp;    }    else {        temp = -INF;        for (int i = 0; i < arr[x].size(); i++){            int y = arr[x][i].v;            dfs(y,cnt+1,pres+arr[x][i].w);            if (pres + dp[y] + arr[x][i].w >= L && pres + dp[y] + arr[x][i].w <= R)                temp = max(temp,dp[y]+arr[x][i].w);                    }        dp[x] = temp;    }}int main(){    while (scanf("%d%d%d",&n,&L,&R) != EOF){        int x,y,z;        for (int i = 0; i < n; i++)            arr[i].clear();        for (int i = 0; i < n-1; i++){            scanf("%d%d%d",&x,&y,&z);            arr[x].push_back(edge(y,z));        }        dfs(0,0,0);        if (dp[0] >= L && dp[0] <= R)            printf("%d\n",dp[0]);        else printf("Oh, my god!\n");    }    return 0;}