Codeforces 453C Little Pony and Summer Sun Celebration(构造)

题目链接：Codeforces 453 Little Pony and Summer Sun Celebration

题目大意：n个节点，m条边，然后m行给定边，最后一行表示每个节点需要进过的次数为奇数次还是偶数次。

解题思路：构造，任意从一个奇数点开始（统一森林的处理），然后每次向下遍历没有经过的节点，并且回溯，每次回溯都要判断一下刚才走过的点满不满足条件，不满足的话就再走一次。最后对于根节点，最后一次回溯要不要走根据当前走过的次数决定。

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;
const int maxn = 1e5+5;

int N, M, a, b, s;
int ans, rec[maxn*4];
int c[maxn], v[maxn];
vector<int> g[maxn];

void init () {
    ans = s = 0;
    scanf("%d%d", &N, &M);
    memset(v, 0, sizeof(v));

    for (int i = 0; i < M; i++) {
        scanf("%d%d", &a, &b);
        g[a].push_back(b);
        g[b].push_back(a);
    }

    for (int i = 1; i <= N; i++) {
        scanf("%d", &c[i]);
        if (c[i])
            s = i;
    }
}

inline void set (int u) {
    c[u] ^= 1;
    rec[ans++] = u;
}

void dfs (int u) {
    set(u);
    v[u] = 1;

    for (int i = 0; i < g[u].size(); i++) {

        if (v[g[u][i]])
            continue;

        dfs(g[u][i]);
        set(u);

        if (c[g[u][i]]) {
            set(g[u][i]);
            set(u);
        }
    }
}

bool judge () {
    for (int i = 1; i <= N; i++)
        if (c[i])
            return false;
    return true;
}

int main () {
    init();
    dfs(s);
    if (c[s]) {
        c[s] = 0;
        ans--;
    }

    if (judge ()) {
        printf("%d\n", ans);
        if (ans) {
            printf("%d", rec[0]);
            for (int i = 1; i < ans; i++)
                printf(" %d", rec[i]);
            printf("\n");
        }
    } else
        printf("-1\n");
    return 0;
}

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