UVA 10054-the Necklace (Euler circuit)

Source: Internet
Author: User

Topic Link: Click to open the link

Test instructions: n beads, the two halves of each bead are made up of different colors. Only the same color can be joined together, asked whether to form a necklace.

Idea: If a bead is regarded as a non-forward edge connecting two vertices, then the subject becomes the existence of the Euler circuit for the non-direction graph.   For the graph, if the number of degrees are even and the graph is connected, then there is a Euler loop. Then start at any point and walk all the way back to the beginning.

See the code for details:

#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include < string> #include <vector> #include <stack> #include <bitset> #include <cstdlib> #include < cmath> #include <set> #include <list> #include <deque> #include <map> #include <queue># Define MAX (a) > (b)? ( A):(B) #define MIN (a) < (b) ( A):(B)) using namespace Std;typedef long Long ll;const double PI = ACOs ( -1.0); const double EPS = 1e-6;const int mod = 10000 00000 + 7;const int INF = 1000000000;const int MAXN = 55;int T,n,m,kase=0,p[maxn],maxv,root,cnt[maxn],g[maxn][maxn];struc T node {int A, B;}        a[1005];void dfs (int u) {for (int v = 1; v <= maxv; v++) if (G[u][v]) {g[u][v]--;        g[v][u]--;        DFS (v);    printf ("%d%d\n", V, U); }}int _find (int x) {return p[x] = = x X:p[x] = _find (P[x]);}    int main () {scanf ("%d", &t);        while (t--) {scanf ("%d", &n);  memset (CNT, 0, sizeof (CNT));      memset (g, 0, sizeof (g));        MAXV = 0;        for (int i=1;i<=50;i++) p[i] = i;        if (Kase) printf ("\ n");            for (int i=0;i<n;i++) {scanf ("%d%d", &a[i].a,&a[i].b);            cnt[a[i].a]++;            cnt[a[i].b]++;            g[a[i].a][a[i].b]++;            g[a[i].b][a[i].a]++;            MAXV = Max (MAXV, Max (A[I].A, a[i].b));            int x = _find (A[I].A);            int y = _find (a[i].b);        if (x! = y) p[x] = y;        } bool OK = true;            for (int i=1;i<=maxv;i++) {if (cnt[i] = = 0) continue; if ((Cnt[i] & 1) | | (_find (i)! = _find (MAXV)))  {OK = false;            Break        }} printf ("Case #%d\n", ++kase);        if (OK) for (int i=1;i<=maxv;i++) {DFS (i);    } else printf ("Some beads may be lost\n"); } return 0;}


UVA 10054-the Necklace (Euler circuit)

Contact Us

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

  • Sales Support

    1 on 1 presale consultation

  • After-Sales Support

    24/7 Technical Support 6 Free Tickets per Quarter Faster Response

  • Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.