[洛谷P1341]无序字母对

题目大意:给一张无向图,找一条字典序最小的欧拉路径

题解:若图不连通或有两个以上的奇数点,则没有欧拉路径,可以$dfs$,在回溯时把这个节点加入答案

卡点:没有在回溯时加入答案,导致出现了欧拉路径没走环(少走了一段)

 

C++ Code:

#include <cstdio>
#include <cctype>
#include <algorithm>
#define maxn 60
int m, start = 52, ind[maxn];
int v[maxn], n, ret[256];
bool e[maxn][maxn];
char ans[maxn * maxn];

int f[maxn];
int find(int x) {return x == f[x] ? x : (f[x] = find(f[x]));}

void dfs(int u) {
	for (int i = 1; i <= n; i++) if (e[u][i]) {
		e[u][i] = e[i][u] = false;
		dfs(i);
	}
	ans[m--] = v[u];
}
int main() {
	scanf("%d", &m);
	for (int i = 'A'; i <= 'Z'; i++) v[++n] = i, ret[i] = n;
	for (int i = 'a'; i <= 'z'; i++) v[++n] = i, ret[i] = n;
	for (int i = 1; i <= n; i++) f[i] = i;
	for (int i = 0; i < m; i++) {
		char ch = getchar();
		while (!isalpha(ch)) ch = getchar();
		int a = ret[static_cast<int> (ch)], b;
		ch = getchar();
		while (!isalpha(ch)) ch = getchar();
		b = ret[static_cast<int> (ch)];
		start = std::min(start, std::min(a, b));
		e[a][b] = e[b][a] = true;
		ind[a]++, ind[b]++;
		f[find(a)] = find(b);
	}
	int cnt = 0;
	for (int i = 1; i <= n; i++) if (ind[i] && f[i] == i) cnt++;
	if (cnt > 1) {
		puts("No Solution");
		return 0;
	}
	cnt = 0;
	for (int i = 1; i <= n; i++) if (ind[i] & 1) {
		if (!cnt) start = i;
		cnt++;
	}
	if (cnt > 2) {
		puts("No Solution");
		return 0;
	}
	
	dfs(start);
	puts(ans);
	return 0;
}

  

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转载自www.cnblogs.com/Memory-of-winter/p/10014705.html