poj 1041 John's trip（欧拉回路入门题）-优快云博客

经典的欧拉回路题

欧拉回路：每条边经过一次且仅一次的称为欧拉回路(euler cycle, euler circuit)。

存在欧拉回路的充要条件：每个点的度数都是偶数, 且图连通。

#include "iostream" #include "stdio.h" #include "string.h" #include "algorithm" using namespace std; int G[50][2000]; //G[点][边] = 点，这样是为了能方便让边lexicographically输出 bool vis[2000]; //记录访问边的情况 int degree[50]; int stack[2000], top; int max(int a, int b) { return a > b ? a : b; } void euler(int cur, int nRoads) //cur当前访问的点 { for(int i = 1; i <= nRoads; i++) { if(!vis[i] && G[cur][i]) //若相邻边未访问过 { vis[i] = true; euler(G[cur][i], nRoads); stack[++top] = i; } } } int main() { int x, y, z, nRoads, nPoints, start; while(scanf("%d%d", &x, &y) && x && y) { nRoads = nPoints = top = 0; memset(vis, false, sizeof(vis)); memset(degree, 0, sizeof(degree)); memset(G, 0, sizeof(G)); start = x < y ? x : y; cin >> z; nRoads = max(nRoads, z); nPoints = max(nPoints, max(x, y)); G[x][z] = y; G[y][z] = x; degree[x]++; degree[y]++; while(scanf("%d%d", &x, &y) && x && y) { cin >> z; nRoads = max(nRoads, z); nPoints = max(nPoints, max(x, y)); G[x][z] = y; G[y][z] = x; degree[x]++; degree[y]++; } bool flag = true; for(int i = 1; i <= nPoints; i++) if(degree[i] & 1) flag = false; if(!flag) cout << "Round trip does not exist." << endl; else { euler(start, nRoads); printf("%d", stack[top]); for(int i = top - 1; i > 0; i--) printf(" %d", stack[i]); cout << endl; } } return 0; }

转载于:https://www.cnblogs.com/firecoder/archive/2011/08/04/2512217.html