HDU 5614 Baby Ming and Matrix tree 树链剖分

题意：

给出一棵树，每个顶点上有个\(2 \times 2\)的矩阵，矩阵有两种操作：

• 顺时针旋转90°，花费是2
• 将一种矩阵替换为另一种矩阵，花费是10

树上有一种操作，将一条路经上的所有矩阵都变为给出的矩阵，并输出最小花费。

分析：

矩阵可以分为两类共6种，一类是两个1相邻的矩阵共4种；一类是两个1在对角线的矩阵共2种。

同一类矩阵可以通过旋转操作得到，否则只能用替换。

事先计算好每种矩阵转换到另外一种矩阵的最少花费，然后树链剖分再用线段树维护就好了。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;const int maxn = 20000 + 10;
const int maxnode = maxn * 4;void read(int& x) {
x = 0;
char c = ' ';
while(c < '0' || c > '9') c = getchar();
while('0' <= c && c <= '9') {
x = x * 10 + c - '0';
c = getchar();
}
}int n, tot;
vector<int> G[maxn];
int son[maxn], sz[maxn], top[maxn], dep[maxn], fa[maxn];
int id[maxn], pos[maxn];void dfs(int u) {
    sz[u] = 1; son[u] = 0;
    for(int v : G[u]) {
        if(v == fa[u]) continue;
        fa[v] = u;
        dep[v] = dep[u] + 1;
        dfs(v);
sz[u] += sz[v];
        if(sz[v] > sz[son[u]]) son[u] = v;
    }
}void dfs2(int u, int tp) {
    id[u] = ++tot;
    pos[tot] = u;
    top[u] = tp;
    if(son[u]) dfs2(son[u], tp);
    for(int v : G[u]) {
        if(v == fa[u] || v == son[u]) continue;
        dfs2(v, v);
    }
}int cost[6][6];
int cntv[maxnode][6], setv[maxnode];int readMat() {
    int a[4];
read(a[0]); read(a[1]); read(a[3]); read(a[2]);
    for(int i = 0; i < 4; i++)
        if(a[i] == 1 && a[(i+1)%4] == 1) return i;
    if(a[0] == 1) return 4;
    return 5;
}int h[maxn];void pushdown(int o, int L, int R) {
    if(setv[o] != -1) {
        int M = (L + R) / 2;
        setv[o<<1] = setv[o<<1|1] = setv[o];
for(int i = 0; i < 6; i++)
cntv[o<<1][i] = cntv[o<<1|1][i] = 0;
        cntv[o<<1][setv[o]] = M - L + 1;
        cntv[o<<1|1][setv[o]] = R - M;
        setv[o] = -1;
    }
}void pushup(int o) {
    for(int i = 0; i < 6; i++)
        cntv[o][i] = cntv[o<<1][i] + cntv[o<<1|1][i];
}void build(int o, int L, int R) {
    if(L == R) { cntv[o][h[pos[L]]] = 1; return; }
    int M = (L + R) / 2;
    build(o<<1, L, M);
    build(o<<1|1, M+1, R);
    pushup(o);
}int q[6];void update(int o, int L, int R, int qL, int qR, int v) {
    if(qL <= L && R <= qR) {
        setv[o] = v;
for(int i = 0; i < 6; i++) q[i] += cntv[o][i];
for(int i = 0; i < 6; i++) cntv[o][i] = 0;
        cntv[o][v] = R - L + 1;
        return;
    }
    pushdown(o, L, R);
    int M = (L + R) / 2;
    if(qL <= M) update(o<<1, L, M, qL, qR, v);
    if(qR > M) update(o<<1|1, M+1, R, qL, qR, v);
    pushup(o);
}void UPDATE(int u, int v, int val) {
memset(q, 0, sizeof(q));
    int t1 = top[u], t2 = top[v];
    while(t1 != t2) {
        if(dep[t1] < dep[t2]) { swap(u, v); swap(t1, t2); }
        update(1, 1, n, id[t1], id[u], val);
        u = fa[t1]; t1 = top[u];
    }
    if(dep[u] > dep[v]) swap(u, v);
    update(1, 1, tot, id[u], id[v], val);
}int main()
{
    for(int i = 0; i < 6; i++)
        for(int j = 0; j < 6; j++) {
if(i == j) { cost[i][j] = 0; continue; }
            int a = ((i >> 2) & 1), b = ((j >> 2) & 1);
if(a ^ b) cost[i][j] = 10;
else if(!a) cost[i][j] = ((((j - i) % 4) + 4) % 4) * 2;
            else cost[i][j] = 2;
        }    int kase; scanf("%d", &kase);
    while(kase--) {
        read(n);
        for(int i = 1; i <= n; i++) G[i].clear();
        for(int i = 1; i < n; i++) {
            int u, v; read(u); read(v);
            G[u].push_back(v);
            G[v].push_back(u);
        }sz[0] = fa[1] = 0;
        dfs(1);        tot = 0;
        dfs2(1, 1);//build segment tree
        memset(cntv, 0, sizeof(cntv));
        memset(setv, -1, sizeof(setv));
        for(int i = 1; i <= n; i++) h[i] = readMat();
        build(1, 1, n);        //Queries
        int _; scanf("%d", &_);
        while(_--) {
            int u, v, val; read(u); read(v);
            val = readMat();
UPDATE(u, v, val);
            int ans = 0;
            for(int i = 0; i < 6; i++) ans += q[i] * cost[i][val];
            printf("%d\n", ans);
        }
    }    return 0;
}