codeforces Round #259(div2) E解题报告

标签：acm   codeforces   dfs   

Twilight Sparkle learnt that the evil Nightmare Moon would return during the upcoming Summer Sun Celebration after one thousand years of imprisonment on the moon. She tried to warn her mentor Princess Celestia, but the princess ignored her and sent her to Ponyville to check on the preparations for the celebration.

Twilight Sparkle wanted to track the path of Nightmare Moon. Unfortunately, she didn‘t know the exact path. What she knew is the parity of the number of times that each place Nightmare Moon visited. Can you help Twilight Sparkle to restore any path that is consistent with this information?

Ponyville can be represented as an undirected graph (vertices are places, edges are roads between places) without self-loops and multi-edges. The path can start and end at any place (also it can be empty). Each place can be visited multiple times. The path must not visit more than 4n places.

3 2
1 2
2 3
1 1 1

3
1 2 3

5 7
1 2
1 3
1 4
1 5
3 4
3 5
4 5
0 1 0 1 0

10
2 1 3 4 5 4 5 4 3 1 

2 0
0 0

0

题目大意：

给出一张图，有N个点，M条边，并给出每个点要求访问次数的奇偶性。，要求输出访问路径。

解法：

首先我们可以明确一点，这就是一个图的遍历，找一个点，设为起点，建立一个搜索遍历树，对于树每一个点，我们完全可以控制奇偶性，假设：

       目前访问的点为v，父节点为fa，如若点v不符合当前的奇偶性，则就让父节点到v绕一次，这样 odd[v] ^= 1, fa[v] ^= 1，这样我们可以完全保证完全控制子节点，将不符合要求的奇偶性调整成符合要求的奇偶性。同时父节点的奇偶性也在改变。

       通过上述的操作，我们可以每次保证子节点的奇偶性符合要求，然而父节点的奇偶性如若不符合要求，则可以通过调整父节点 与 父节点的父节点来调整奇偶性，这样我们就可以奇偶性传递，最终传递到根节点。根节点如若不符合该如何调整呢？由于我们是走遍历树，到达叶节点还要回来的，意味着我们要走至少两次根节点，一次是出发，一次是最后一次回归。我们可以将根节点 r1 调整到根节点的其中一个子节点r2，使得奇偶性再次得到调整

代码：

#include <cstdio>
#include <vector>
#define N_max 123456

using namespace std;

int n, m, fa[N_max], want[N_max];
int Odd_n, Odd_x, haveOdd[N_max];
vector <int> G[N_max], ans;

int getf(int x) {
	return (fa[x] == x) ? x : fa[x] = getf(fa[x]);
}
void addedge(int x, int y) {
	G[x].push_back(y);
}

void init() {
	scanf("%d%d", &n, &m);

	for (int i = 1; i <= n; i++)  fa[i] = i;
	for (int i = 1; i <= m; i++) {
		int x, y;

		scanf("%d%d", &x, &y);

		int tmpx=getf(x);
		int tmpy=getf(y);
		if (tmpx != tmpy) {
			fa[tmpx] = tmpy;
			addedge(x, y);
			addedge(y, x);
		}
	}

	for (int i = 1; i <= n; i++) {
		scanf("%d", &want[i]);

		if (want[i])  haveOdd[getf(i)] = 1;
	}

	for (int i = 1; i <= n; i++)
		if (haveOdd[i]) {
			Odd_n++;
			Odd_x = i;
		}
}

void dfs(int now, int pre) {
	ans.push_back(now);
	want[now] ^= 1;

	for (int i = 0; i < G[now].size(); i++) {
		int nex = G[now][i];
		if (nex == pre)  continue;

		dfs(nex, now);
		ans.push_back(now);
		want[now] ^= 1;

		if (want[nex]) {
			ans.push_back(nex);
			want[nex] ^= 1;
			ans.push_back(now);
			want[now] ^= 1;
		}
	}
}

void solve() {
	if (Odd_n == 0) {
		printf("0\n");
		return;
	}

	if (Odd_n > 1) {
		printf("-1\n");
		return;
	}

	dfs(Odd_x, -1);
	int x = 0;
	if (want[Odd_x])  x=1;

	printf("%d\n", ans.size() - x);
	for (int i = x; i < ans.size(); i++)
		printf("%d ", ans[i]);
}

int main() {
	init();
	solve();
}

codeforces Round #259(div2) E解题报告

标签：acm   codeforces   dfs   

原文地址：http://blog.csdn.net/lmhunter/article/details/40432301