• John's trip(POJ1041+欧拉回路+打印路径)


    题目链接:http://poj.org/problem?id=1041

    题目:

    题意:给你n条街道,m个路口,每次输入以0 0结束,给你的u v t分别表示路口u和v由t这条街道连接,要输出从起点出发又回到起点的字典序最小的路径,如果达不到输出Round trip does not exist.

    思路:首先得判断是否存在欧拉回路,如果不存在则输出“Round trip does not exist.”。记录每个路口的度,如果存在度为奇数得路口则是不存在欧拉回路得图,否则用mp[u][t]=v来表示u可以通过t这条街道到达v,跑一边欧拉回路并记录路径即可。

    代码实现如下:

     1 #include <set>
     2 #include <map>
     3 #include <queue>
     4 #include <stack>
     5 #include <cmath>
     6 #include <bitset>
     7 #include <cstdio>
     8 #include <string>
     9 #include <vector>
    10 #include <cstdlib>
    11 #include <cstring>
    12 #include <iostream>
    13 #include <algorithm>
    14 using namespace std;
    15 
    16 typedef long long ll;
    17 typedef pair<ll, ll> pll;
    18 typedef pair<ll, int> pli;
    19 typedef pair<int, ll> pil;;
    20 typedef pair<int, int> pii;
    21 typedef unsigned long long ull;
    22 
    23 #define lson i<<1
    24 #define rson i<<1|1
    25 #define bug printf("*********
    ");
    26 #define FIN freopen("D://code//in.txt", "r", stdin);
    27 #define debug(x) cout<<"["<<x<<"]" <<endl;
    28 #define IO ios::sync_with_stdio(false),cin.tie(0);
    29 
    30 const double eps = 1e-8;
    31 const int mod = 10007;
    32 const int maxn = 1e6 + 7;
    33 const double pi = acos(-1);
    34 const int inf = 0x3f3f3f3f;
    35 const ll INF = 0x3f3f3f3f3f3f3f;
    36 
    37 int s, u, v, t, mx, p;
    38 int mp[55][2007], in[55], vis[2007], ans[2007];
    39 
    40 void eulergraph(int s) {
    41     for(int i = 1; i <= mx; i++) {
    42         if(mp[s][i] && !vis[i]) {
    43             vis[i] = 1;
    44             eulergraph(mp[s][i]);
    45             ans[++p] = i;
    46         }
    47     }
    48 }
    49 
    50 int main() {
    51     //FIN;
    52     while(~scanf("%d%d", &u, &v)) {
    53         if(u == 0 && v == 0) break;
    54         s = min(u, v);
    55         p = 0;
    56         memset(in, 0, sizeof(vis));
    57         memset(mp, 0, sizeof(mp));
    58         memset(vis, 0, sizeof(vis));
    59         scanf("%d", &t);
    60         in[u]++, in[v]++;
    61         mx = t;
    62         mp[u][t] = v, mp[v][t] = u;
    63         while(~scanf("%d%d", &u, &v)) {
    64             if(u == 0 && v == 0) break;
    65             scanf("%d", &t);
    66             mx = max(mx, t);
    67             in[u]++, in[v]++;
    68             mp[u][t] = v, mp[v][t] = u;
    69         }
    70         int flag = 0;
    71         for(int i = 1; i <= 45; i++) {
    72             if(in[i] & 1) {
    73                 printf("Round trip does not exist.
    ");
    74                 flag = 1;
    75                 break;
    76             }
    77         }
    78         if(flag) continue;
    79         eulergraph(s);
    80         for(int i = p; i >= 1; i--) {
    81             printf("%d%c", ans[i], i == 1 ? '
    ' : ' ');
    82         }
    83     }
    84     return 0;
    85 }
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  • 原文地址:https://www.cnblogs.com/Dillonh/p/9396388.html
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