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@ -8,26 +8,31 @@ void add(int a, int b, int c = 0) {
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e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
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}
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int n, c[N], f1[N], f2[N];
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int n, c[N];
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// 设白点个数cnt1,黑点个数cnt2, f[u]:以u为根的子树中,cnt1-cnt2的最大值
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int f1[N], f2[N];
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void dfs1(int u, int fa) {
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if (c[u])
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f1[u] = 1;
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f1[u] = 1; // 如果u是白色则cnt1=1,那么目前黑色数量cnt2=0,cnt1-cnt2是固定值=1
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else
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f1[u] = -1;
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f1[u] = -1; // 如果u是黑色则cnt1=0,那么目前黑色数量cnt2=1,cnt1-cnt2是固定值=0-1=-1
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for (int i = h[u]; ~i; i = ne[i]) {
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int v = e[i];
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if (v == fa) continue;
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dfs1(v, u);
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f1[u] += max(f1[v], 0);
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f1[u] += max(f1[v], 0); // 为了贪图cnt1-cnt2的最大值,那么对结果有贡献的就竞争一下,如果都小于0了,就退出评比
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}
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}
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// 换根dp
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void dfs2(int u, int fa) {
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for (int i = h[u]; ~i; i = ne[i]) {
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int v = e[i];
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if (v == fa) continue;
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// 本题的核心:递推式
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f2[v] = max(f2[u] + f1[u] - max(f1[v], 0), 0);
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dfs2(v, u);
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}
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@ -37,18 +42,18 @@ int main() {
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memset(h, -1, sizeof h);
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cin >> n;
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for (int i = 1; i <= n; i++) cin >> c[i];
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for (int i = 1; i <= n; i++) cin >> c[i]; // 每个节点都有自己的颜色c[i]
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for (int i = 1; i <= n - 1; i++) {
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for (int i = 1; i < n; i++) { // n-1条边
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int a, b;
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cin >> a >> b;
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add(a, b), add(b, a);
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}
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dfs1(1, 1);
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dfs2(1, 1);
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// 第一次dfs
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dfs1(1, 0);
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// 第二次dfs,换根
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dfs2(1, 0);
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// 输出
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for (int i = 1; i <= n; i++) printf("%d ", f1[i] + f2[i]);
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return 0;
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}
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