zoukankan      html  css  js  c++  java
  • poj 3255 Roadblocks 次短路(两次dijksta)

    Roadblocks

    Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)
    Total Submission(s) : 15   Accepted Submission(s) : 6
    Problem Description

    Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too quickly, because she likes the scenery along the way. She has decided to take the second-shortest rather than the shortest path. She knows there must be some second-shortest path.

    The countryside consists of R (1 ≤ R ≤ 100,000) bidirectional roads, each linking two of the N (1 ≤ N ≤ 5000) intersections, conveniently numbered 1..N. Bessie starts at intersection 1, and her friend (the destination) is at intersection N.

    The second-shortest path may share roads with any of the shortest paths, and it may backtrack i.e., use the same road or intersection more than once. The second-shortest path is the shortest path whose length is longer than the shortest path(s) (i.e., if two or more shortest paths exist, the second-shortest path is the one whose length is longer than those but no longer than any other path).

     
    Input
    Line 1: Two space-separated integers: <i>N</i> and <i>R</i> <br>Lines 2..<i>R</i>+1: Each line contains three space-separated integers: <i>A</i>, <i>B</i>, and <i>D</i> that describe a road that connects intersections <i>A</i> and <i>B</i> and has length <i>D</i> (1 ≤ <i>D</i> ≤ 5000)
     
    Output
    Line 1: The length of the second shortest path between node 1 and node <i>N</i>
     
    Sample Input
    4 4 1 2 100 2 4 200 2 3 250 3 4 100
     
    Sample Output
    450
     
    有n个路口r条马路,马路可以重复走,问从1号路口到n号路口的次短路
     

    次短路问题,实际上可以这么理解:

    在知道最短路的情况下,不走最短路,绕一段路,而且只能绕一段路,否则会不满足次短。

    所以就先找到最短路并记录下路径,然后枚举最短路上的每一个点a,从这个点再绕一个点b,然后再加上点b到n的最短路。

    所以我们需要知道从1到每个点的最短路,还需要知道从每个点到n的最短路,从每个点到n的最短路就是从n到每个点的最短路

    所以两次dijkstra 然后枚举次短路就好啦

     

    邻接矩阵居然超内存

     1 #include <cstring>  
     2 #include <cstdio>  
     3 #include <iostream>  
     4 #include <vector> 
     5 #include <algorithm>
     6 #include <string>
     7 #define inf 0x3f3f3f3f 
     8 const int maxn = 5010;
     9 using namespace std;
    10 int n, m;
    11 int d1[maxn];
    12 int d2[maxn];
    13 bool book[maxn];
    14 struct edge
    15 {
    16     int to, c;
    17 };
    18 vector<edge> e[maxn];
    19 //用邻接矩阵会超内存
    20 void dijkstra(int s, int *d)
    21 {
    22     memset(d,inf, maxn * sizeof(int));
    23     memset(book, 0, sizeof(book));
    24     int i;
    25     //for (i = 0; i < e[s].size(); i++) d[e[s][i].to] = e[s][i].c;
    26     //book[s] = 1;
    27     //不能直接赋值,也要进行比较,典型样例
    28     //2 2
    29     //1 2 100
    30     //1 2 200
    31     d[s] = 0;
    32     while (1)
    33     {
    34         int k = -1;
    35 
    36         int min = inf;
    37         for ( i = 1; i<= n;i++)
    38         {
    39             if (!book[i] && d[i] < min)
    40             {
    41                 min = d[i];
    42                 k = i;
    43             }
    44         }
    45         if (k == -1) break;
    46 
    47         else
    48         {
    49             book[k] = 1;
    50             for (i=0;i<e[k].size();i++)
    51             {
    52                 if (d[e[k][i].to] > d[k] + e[k][i].c)
    53                 {
    54                     d[e[k][i].to] = d[k] + e[k][i].c;
    55                 }
    56 
    57             }
    58         }
    59     }
    60 }
    61 
    62 int main()
    63 {
    64     int i;
    65     cin >> n >> m;
    66     for (i = 1; i <= m; i++)
    67     {
    68         edge t, t1;
    69         int k;
    70         cin >> k >> t.to >> t.c;
    71         t1.to = k;
    72         t1.c = t.c;
    73         e[k].push_back(t);//双向存。存一个点出发的多个目的地从k出发,目的地是t.to,花费t.c
    74         e[t.to].push_back(t1);
    75     }
    76 
    77     dijkstra(1, d1);
    78     dijkstra(n, d2);//某个点到n的最短路就是n到某个点的最短路
    79     int k = n;
    80     int ans = inf;
    81     int minn = d1[n];
    82 
    83     for (k=1;k<=n;k++)
    84     {
    85         for (i = 0; i<e[k].size(); i++)
    86         {
    87             int ee = d1[k] + e[k][i].c + d2[e[k][i].to];
    88             if (ans>ee&&ee>minn)
    89             {
    90                 ans = ee;
    91             }
    92         }
    93     }
    94     cout << ans << endl;
    95     return 0;
    96 }

    超内存代码且错误代码

     1 #include <iostream>
     2 #include <cstring>
     3 #include <string>
     4 #include <algorithm>
     5 #include <cstdio>
     6 #define inf 0x3f3f3f3f
     7 using namespace std;
     8 int e[5005][5005];
     9 int d1[1005];
    10 int d2[1005];
    11 int m, n;
    12 void dijkstra(int s, int *d)
    13 {
    14     int book[1005];
    15     memset(book, 0, sizeof(book));
    16     int i;
    17     for (i = 1; i <= n; i++)
    18     {
    19         d[i] = e[s][i];
    20     }
    21     book[1] = 1;
    22     while (1)
    23     {
    24         int min = inf;
    25         int k = -1;
    26         for (i = 1; i <= n; i++)
    27         {
    28             if (d[i] < min&&book[i]==0)
    29             {
    30                 min = d[i];
    31                 k = i;
    32             }
    33         }
    34         if (k == -1) break;
    35         book[k] = 1;
    36         for (i = 1; i <= n; i++)
    37         {
    38             if (book[i] == 0 && d[i] > d[k] + e[k][i])
    39             {
    40                 d[i] = d[k] + e[k][i];
    41             }
    42         }
    43     }
    44 }
    45 int main()
    46 {
    47     int i, j;
    48     scanf("%d %d", &m, &n);
    49     for (i = 1; i <= n; i++)
    50     {
    51         for (j = 1; j <= n; j++)
    52         {
    53             if (i == j) e[i][j] = 0;
    54             else
    55                 e[i][j] = inf;
    56         }
    57     }
    58     for (i = 1; i <= n; i++)
    59     {
    60         int x, y, z;
    61         scanf("%d %d %d", &x, &y,&z);
    62         if (e[x][y] > z)
    63         {
    64             e[x][y] = z;
    65             e[y][x] = z;
    66         }
    67     }
    68     dijkstra(1, d1);
    69     dijkstra(n, d2);
    70     int minn = d1[n];
    71     int ans = inf;
    72     for (i = 1; i <= n; i++)
    73     {
    74         for (j = 1; j <= n; j++)
    75         {
    76             if (i == j) continue;
    77             if (e[i][j] == inf) continue;
    78             if (d1[i] + e[i][j] + d2[j] > minn)
    79             {
    80                 ans = min(ans, d1[i] + e[i][j] + d2[j]);
    81             }
    82         }
    83     }
    84     printf("%d
    ", ans);
    85     return 0;
    86 }
    View Code
  • 相关阅读:
    对称加密和非对称加密
    数字签名
    内存溢出和内存泄漏
    生产随机字符串
    天才高中生参与斯坦福新研究:在图像压缩上,人类比算法强
    用机器人替代宇航员!日本打造远程操作机器人
    机器人也能拥有“物理直觉”?谷歌这款机器人真是厉害了
    35所高校新设人工智能本科专业 专家:人工智能非孤立专业
    “入职1年,我和做 AI 的同学薪资翻了 2 翻!”
    让机器学会认脸究竟有多少用处
  • 原文地址:https://www.cnblogs.com/caiyishuai/p/8438600.html
Copyright © 2011-2022 走看看