Description
给定一棵树，求它的最小点权覆盖集，其中允许强制某个点选或不选
Solution
ddp用LCT维护
明确一个关系式：最小点权覆盖集=全集-最大点权独立集
那么n≤2000的暴力就很简单了，暴力修改，然后求最大点权独立集就好了（我在考场上就是这么写的）
关于正解，我采用的是动态dp（动态dp的题解点这里）
根据上面的式子，我们用动态dp维护最大点权独立集，然后用全集减去就好了
然后我们考虑强制选点的限制
如果强制一个点选，那么我们将这个点的点权加上负无穷
如果强制不选，那么我们将这个点的点权加上正无穷，然后全集加上正无穷就好了
记得每次操作后要还原现场
时间复杂度是$O(mlogn)$
Code
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int N = 1e5 + 10; 5 const ll INF = 1ll << 40; 6 inline int read() { 7 int ret = 0, op = 1; 8 char c = getchar(); 9 while (!isdigit(c)) { 10 if (c == '-') op = -1; 11 c = getchar(); 12 } 13 while (isdigit(c)) { 14 ret = (ret << 3) + (ret << 1) + c - '0'; 15 c = getchar(); 16 } 17 return ret * op; 18 } 19 struct Edge { 20 int nxt, to; 21 } e[N << 1]; 22 int num, head[N]; 23 struct Matrix { 24 ll m[2][2]; 25 Matrix() { 26 m[0][0] = m[0][1] = m[1][0] = m[1][1] = -INF; 27 } 28 inline ll getans() { 29 return max(m[0][0], m[1][0]); 30 } 31 inline void build(ll x, ll y) { 32 m[0][0] = m[0][1] = x; 33 m[1][0] = y; m[1][1] = -INF; 34 } 35 Matrix operator *(const Matrix &x) const { 36 Matrix ret; 37 for (register int i = 0; i < 2; ++i) 38 for (register int j = 0; j < 2; ++j) 39 for (register int k = 0; k < 2; ++k) 40 ret.m[i][j] = max(ret.m[i][j], m[i][k] + x.m[k][j]); 41 return ret; 42 } 43 }; 44 struct LCT { 45 int fa, ch[2]; 46 ll f[2]; 47 Matrix x; 48 } a[N]; 49 int n, m, val[N]; 50 ll sum; 51 string s; 52 inline void add(int from, int to) { 53 e[++num].nxt = head[from]; 54 e[num].to = to; 55 head[from] = num; 56 } 57 void dfs(int now, int fa) { 58 a[now].f[1] = val[now]; 59 for (register int i = head[now]; i; i = e[i].nxt) { 60 int to = e[i].to; 61 if (to == fa) continue ; 62 a[to].fa = now; 63 dfs(to, now); 64 a[now].f[0] += max(a[to].f[1], a[to].f[0]); 65 a[now].f[1] += a[to].f[0]; 66 } 67 a[now].x.build(a[now].f[0], a[now].f[1]); 68 } 69 inline void update(int now) { 70 a[now].x.build(a[now].f[0], a[now].f[1]); 71 if (a[now].ch[0]) a[now].x = a[a[now].ch[0]].x * a[now].x; 72 if (a[now].ch[1]) a[now].x = a[now].x * a[a[now].ch[1]].x; 73 } 74 inline int isnroot(int now) { 75 return a[a[now].fa].ch[0] == now || a[a[now].fa].ch[1] == now; 76 } 77 inline void rotate(int x) { 78 int y = a[x].fa; 79 int z = a[y].fa; 80 int xson = a[y].ch[1] == x; 81 int yson = a[z].ch[1] == y; 82 int B = a[x].ch[xson ^ 1]; 83 if (isnroot(y)) a[z].ch[yson] = x; 84 a[y].ch[xson] = B; 85 a[x].ch[xson ^ 1] = y; 86 if (B) a[B].fa = y; 87 a[y].fa = x; 88 a[x].fa = z; 89 update(y); update(x); 90 } 91 void splay(int x) { 92 while (isnroot(x)) { 93 int y = a[x].fa; 94 int z = a[y].fa; 95 if (isnroot(y)) 96 (a[y].ch[0] == x) ^ (a[z].ch[0] == y) ? rotate(x) : rotate(y); 97 rotate(x); 98 } 99 update(x); 100 } 101 void access(int x) { 102 for (register int y = 0; x; y = x, x = a[x].fa) { 103 splay(x); 104 if (a[x].ch[1]) { 105 a[x].f[0] += a[a[x].ch[1]].x.getans(); 106 a[x].f[1] += a[a[x].ch[1]].x.m[0][0]; 107 } 108 if (y) { 109 a[x].f[0] -= a[y].x.getans(); 110 a[x].f[1] -= a[y].x.m[0][0]; 111 } 112 a[x].ch[1] = y; 113 update(x); 114 } 115 } 116 void change(int x, ll y) { 117 access(x); splay(x); 118 a[x].f[1] += y; 119 update(x); 120 } 121 void solve(int x, int op1, int y, int op2) { 122 change(x, op1 ? -INF : INF); 123 change(y, op2 ? -INF : INF); 124 splay(1); 125 sum += ((op1 ^ 1) + (op2 ^ 1)) * INF; 126 ll ans = sum - a[1].x.getans(); 127 change(x, op1 ? INF : -INF); 128 change(y, op2 ? INF : -INF); 129 sum -= ((op1 ^ 1) + (op2 ^ 1)) * INF; 130 if (ans > INF) puts("-1"); 131 else printf("%lld\n", ans); 132 } 133 int main() { 134 n = read(); m = read(); cin >> s; 135 for (register int i = 1; i <= n; ++i) { 136 val[i] = read(); 137 sum += val[i]; 138 } 139 for (register int i = 1; i < n; ++i) { 140 int x = read(), y = read(); 141 add(x, y); add(y, x); 142 } 143 dfs(1, 0); 144 while (m--) { 145 int x = read(), op1 = read(), y = read(), op2 = read(); 146 solve(x, op1, y, op2); 147 } 148 return 0; 149 }