zoukankan      html  css  js  c++  java
  • POJ1985Cow Marathon[树的直径]

    Cow Marathon
    Time Limit: 2000MS   Memory Limit: 30000K
    Total Submissions: 5117   Accepted: 2492
    Case Time Limit: 1000MS

    Description

    After hearing about the epidemic of obesity in the USA, Farmer John wants his cows to get more exercise, so he has committed to create a bovine marathon for his cows to run. The marathon route will include a pair of farms and a path comprised of a sequence of roads between them. Since FJ wants the cows to get as much exercise as possible he wants to find the two farms on his map that are the farthest apart from each other (distance being measured in terms of total length of road on the path between the two farms). Help him determine the distances between this farthest pair of farms. 

    Input

    * Lines 1.....: Same input format as "Navigation Nightmare".

    Output

    * Line 1: An integer giving the distance between the farthest pair of farms. 

    Sample Input

    7 6
    1 6 13 E
    6 3 9 E
    3 5 7 S
    4 1 3 N
    2 4 20 W
    4 7 2 S
    

    Sample Output

    52
    

    Hint

    The longest marathon runs from farm 2 via roads 4, 1, 6 and 3 to farm 5 and is of length 20+3+13+9+7=52. 

    Source


    裸题
    换一种DP写法
    小心不能用f[j][1]来更新f[i],因为是从i节点的不同子树中选
    //
    //  main.cpp
    //  poj1985
    //
    //  Created by Candy on 9/21/16.
    //  Copyright ? 2016 Candy. All rights reserved.
    //
    
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    const int N=5e4+5;
    int read(){
        char c=getchar();int x=0,f=1;
        while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
        while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
        return x*f;
    }
    int n,m,u,v,w,ans=0; char c[2];
    struct edge{
        int ne,v,w;
    }e[N<<1];
    int h[N],cnt=0;
    void ins(int u,int v,int w){
        cnt++;
        e[cnt].v=v; e[cnt].w=w; e[cnt].ne=h[u]; h[u]=cnt;
        cnt++;
        e[cnt].v=u; e[cnt].w=w; e[cnt].ne=h[v]; h[v]=cnt;
    }
    int f[N][2];
    void dp(int u,int fa){
        f[u][0]=0;
        for(int i=h[u];i;i=e[i].ne){
            int v=e[i].v,w=e[i].w;
            if(v==fa) continue;
            dp(v,u);
            if(f[v][0]+w>f[u][0]){
                f[u][1]=f[u][0];
                f[u][0]=f[v][0]+w;
            }else{
                if(f[v][0]+w>f[u][1]) f[u][1]=f[v][0]+w;
            }
        }
    }
    int main(int argc, const char * argv[]) {
        n=read();m=read();
        for(int i=1;i<=m;i++){
            u=read();v=read();w=read();
            ins(u,v,w);
            scanf("%s",c);
        }
        memset(f,-1,sizeof(f));
        dp(1,0);
        for(int i=1;i<=n;i++) ans=max(ans,f[i][1]+f[i][0]);
        printf("%d",ans);
        return 0;
    }
  • 相关阅读:
    1. Go的安装和第一行代码
    合工大OJ 1344
    __int64与long long、long的区别
    合工大OJ 1343
    如何快速查找下载java项目所需jar包
    油田勘测(深度优先算法,广度优先算法)
    图的创建(邻接矩阵)
    五大常用算法总结
    前序遍历,中序遍历,后序遍历(树的深度优先算法),层序遍历(树的广度优先算法)
    IE CSS Hack
  • 原文地址:https://www.cnblogs.com/candy99/p/5893366.html
Copyright © 2011-2022 走看看