poj 3268 Silver Cow Party

最新推荐文章于 2021-04-01 18:48:34 发布

转载最新推荐文章于 2021-04-01 18:48:34 发布 · 46 阅读

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原文链接：http://www.cnblogs.com/smilesundream/p/6642535.html

本博客探讨了在给定农场之间的单向道路网络中，如何通过Dijkstra算法解决牛群从各自农场到聚会地点及返回时的最长时间行走问题。详细介绍了算法的应用、时间复杂度分析和优化策略。

Silver Cow Party

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 16277		Accepted: 7450

Description

One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X(1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires T_i (1 ≤ T_i ≤ 100) units of time to traverse.

Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow's return route might be different from her original route to the party since roads are one-way.

Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?

Input

Line 1: Three space-separated integers, respectively: N, M, and X
Lines 2..M+1: Line i+1 describes road i with three space-separated integers: A_i, B_i, and T_i. The described road runs from farm A_i to farm B_i, requiring T_i time units to traverse.

Output

Line 1: One integer: the maximum of time any one cow must walk.

Sample Input

Sample Output

Hint

Cow 4 proceeds directly to the party (3 units) and returns via farms 1 and 3 (7 units), for a total of 10 time units.

Source

USACO 2007 February Silver

错因：错在没有正确的了解时间复杂度吧，看到n==10^3，想到warshall算法的复杂度是

n^3，算出来为10^9然后看到2s的时间限制就觉得不会超时，后来才知道一般10^8的复杂度才能算1s，就只能换dijkstra了，这样的使用邻接矩阵的复杂度是n^2，就不会超时了

分析：使用djjkstra的话需要扫两遍，第一次扫了后将边方向全部反过来再扫一遍，关键是将边取反这个操作，写的不好的话又会很复杂，学习了一个新的算法，好好体会下

<span style="font-size:18px;">void tra()
{
    int temp;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=i;j++)  //注意是就j<=i
           {
               temp=map[i][j];
               map[i][j]=map[j][i];
               map[j][i]=temp;
           }
}           </span>

下面是总的代码：

<span style="font-size:18px;">#include<cstdio>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 1e3+5;
const int mod = 1000000007;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
int map[1005][1005],dist[1005],used[1005],ans[1005];
int n,m,s;
void init()
{
    memset(map,inf,sizeof(map));
    memset(ans,0,sizeof(ans));
    for(int i=1;i<=n;i++)
        map[i][i]=0;
}
int min(int a ,int b)
{
    return a>b?b:a;
}
void tra()
{
    int temp;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=i;j++)
           {
               temp=map[i][j];
               map[i][j]=map[j][i];
               map[j][i]=temp;
           }
}                             //数组反向
void dijkstra()
{
    memset(used,0,sizeof(used));
    memset(dist,inf,sizeof(dist));
    dist[s]=0;
    while(1)
    {
    int u=0,minn=inf;
    for(int i=1;i<=n;i++)
        if(!used[i]&&dist[i]<minn)
        {
            u=i;
            minn=dist[i];
        }
    if(u==0)
        break;
    used[u]=1;
    for(int i=1;i<=n;i++)
        if(!used[i]&&dist[i]>dist[u]+map[u][i])
            dist[i]=dist[u]+map[u][i];
    }
    for(int i=1;i<=n;i++)
        ans[i]+=dist[i];
}
int main()
{
    while(~scanf("%d %d %d",&n,&m,&s))
    {
         init();
         int x,y,c;
         for(int i=1;i<=m;i++)
            {
                scanf("%d %d %d",&x,&y,&c);
                map[x][y]=c;
            }
        dijkstra();
        tra();
        dijkstra();
        int maxn=0;
        for(int i=1;i<=n;i++)
            if(ans[i]>maxn)
            maxn=ans[i];
        printf("%d\n",maxn);
    }
    return 0;
}
</span>

转载于:https://www.cnblogs.com/smilesundream/p/6642535.html