题意：
考虑一定是属于\(a\)的在一坨，属于\(b\)的在一坨，找到这条连接\(a\)和\(b\)的边，然后分别直接按\(dfs\)序染色即可 注意属于\(a\)的连通块或属于\(b\)的连通块可能在\(dfs\)树上不都体现为一棵完整的子树，所以需要都判断一下
#include<bits/stdc++.h> #define N (200000 + 10) using namespace std; inline int read() { int cnt = 0, f = 1; char c = getchar(); while (!isdigit(c)) {if (c == '-') f = -f; c = getchar();} while (isdigit(c)) {cnt = (cnt << 3) + (cnt << 1) + (c ^ 48); c = getchar();} return cnt * f; } int n, a, b, x, y, fa[N], siz[N], id[N], val; int first[N], to[N], nxt[N], tot; void add (int x, int y) {nxt[++tot] = first[x], first[x] = tot, to[tot] = y;} void get_siz(int x, int father) { siz[x] = 1; fa[x] = father; for (register int i = first[x]; i; i = nxt[i]) { int v = to[i]; if (v == father) continue; get_siz(v, x), siz[x] += siz[v]; } } void print(int x, int fa, int d) { for (register int i = first[x]; i; i = nxt[i]) { int v = to[i]; if (v == fa) continue; print(v, x, d); } val += d; id[x] = val; } void work () { bool ok = 0; get_siz(1, 0); for (register int i = 1; i <= n; ++i) if (siz[i] == a) { val = 0; print(i, fa[i], 1); val = 0; print(fa[i], i, -1); ok = 1; } else if (siz[i] == b) { val = 0; print(i, fa[i], -1); val = 0; print(fa[i], i, 1); ok = 1; } if (!ok) puts("-1"); else for (register int i = 1; i <= n; ++i) printf("%d ", id[i]); putchar('\n'); } int main() { n = read(), a = read(), b = read(); for (register int i = 1; i <= n - 1; ++i) { x = read(), y = read(); add(x, y), add(y, x); } work(); return 0; }首先能发现一个性质，括号结构最多嵌套两层 如果有三层的嵌套可以这样化简 ( ( ( ) ) ) ↓ ( ( ) ( ) ) 符号反三层和反一层等价 设\(f[i][0~2]\)表示当前处理到第\(i\)位，前面有\(0/1/2\)个未配对的左括号 在当前数\(>0\)时状态转移考虑补右括号，\(<0\)时考虑先强行在“-”后补一个左括号（如果对答案不优，自然会在后面的更新中去掉：-(a) + b），再进行转移 具体方程看代码
#include<bits/stdc++.h> #define int long long using namespace std; inline int read() { int cnt = 0, f= 1;char c = getchar(); while (!isdigit(c)) {if (c == '-') f = -f; c = getchar();} while (isdigit(c)) {cnt = (cnt << 3) + (cnt << 1) + (c ^ 48); c = getchar();} return cnt * f; } int T, n, a[500010], dp[500010][3]; signed main() { T = read(); while (T--) { n = read(); for (register int i = 1; i <= n; ++i) a[i] = read(); dp[0][0] = 0, dp[0][1] = dp[0][2] = -1e18; for (register int i = 1; i <= n; ++i) if (a[i] > 0) { dp[i][0] = max(max(dp[i - 1][0] + a[i], dp[i - 1][1] + a[i]), dp[i - 1][2] + a[i]); dp[i][1] = max(dp[i - 1][1] - a[i], dp[i - 1][2] - a[i]); dp[i][2] = dp[i - 1][2] + a[i]; } else { dp[i][0] = -1e18; dp[i][1] = max(max(dp[i - 1][0] + a[i], dp[i - 1][1] + a[i]), dp[i - 1][2] + a[i]); dp[i][2] = max(dp[i - 1][1] - a[i], dp[i - 1][2] - a[i]); } printf("%lld\n", max(max(dp[n][0], dp[n][1]), dp[n][2])); } return 0; }T3太麻烦了不想写，先咕了
