poj 1041 John's trip (Euler Circuit)

1041 -- John's trip

　　输出字典序最小的欧拉回路。之前做过的是要求点的序列最小字典序，今天见到这个经典的欧拉回路是要输出边的编号的最小序。做法跟输出点路径的相似。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <stack>

using namespace std;

typedef pair<int, int> PII;
typedef multiset<PII> MTPII;
#define ITOR iterator
#define MPR make_pair
#define FI first
#define SE second
#define REP(i, n) for (int i = 0; i < (n); i++)
const int N = 50;
MTPII rel[N];
map<int, int> nextVex[N];
int roadCnt;

int input() {
    roadCnt = 0;
    int x, y, z, mk = 0;
    REP(i, N) rel[i].clear(), nextVex[i].clear();;
    while (cin >> x >> y && (x || y)) {
        cin >> z;
        if (!roadCnt) {
            mk = min(x, y);
        }
        rel[x].insert(MPR(z, y));
        nextVex[x][z] = y;
        rel[y].insert(MPR(z, x));
        nextVex[y][z] = x;
        roadCnt++;
    }
    return mk;
}

stack<int> path;

int check(int mk) {
    REP(i, N) if (rel[i].size() & 1) return -1;
    stack<int> tmp = path;
    int sz = tmp.size(), t;
    REP(i, sz) {
        t = tmp.top();
        tmp.pop();
        if (nextVex[mk].find(t) == nextVex[mk].end()) return -1;
        mk = nextVex[mk][t];
    }
    return mk;
}

void dfs(int x) {
    while (!rel[x].empty()) {
        int edge = (*rel[x].begin()).FI, nx = (*rel[x].begin()).SE;
        rel[x].erase(rel[x].begin());
        rel[nx].erase(MPR(edge, x));
        dfs(nx);
        path.push(edge);
    }
}

void output() {
    int sz = path.size();
    REP(i, sz) {
        if (i) putchar(' ');
        printf("%d", path.top());
        path.pop();
    }
    puts("");
}

int main() {
//    freopen("in", "r", stdin);
    int mk;
    while (mk = input()) {
        while (!path.empty()) path.pop();
        dfs(mk);
        if ((path.size() == roadCnt) && (check(mk) == mk)) output();
        else puts("Round trip does not exist.");
    }
    return 0;
}

——written by Lyon