poj 1041 John's trip（欧拉回路入门题）

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经典的欧拉回路题

http://poj.org/problem?id=1041

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;}