bzoj 2286: [Sdoi2011]消耗战

题目链接

bzoj 2286: [Sdoi2011]消耗战

题解

抽离虚树dp
对于虚树我们可以O(m)构造
dfs序排序后，易证相邻两点lca为所有出现到的lca
每次维护一个深度递增的栈，用其来维护一个节点的虚树，当其被pop出栈时他的虚数也构造完了
那么栈中序列即为链的父子关系
每次若将要插入一点与栈顶点的lca，位于栈顶上个点的虚树树中，且不为栈顶点的后代<若是的话直接压入栈中，继续跟新该链>，那么弹出栈顶点后直接压入栈中，连边，否则pop到是为止，此时开始新链更新
pop时连边建立父子关系就好了
然后就是普通的树形dp
ps:我一定要出一道虚树 ~~毒瘤~~ 码农题

代码

#include<cstdio>
#include<algorithm>
#include<cstring> 
#define LL long long 
inline int read() {
    int x = 0,f =  1;
    char c = getchar();
    while(c < '0' || c > '9') {if(c == '-')f = -1;c = getchar();} 
    while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = getchar(); 
    return x * f;
} 
const int maxn = 2000007;  
struct node {
    int w,v,next;
} edge[maxn << 1],e[maxn << 1]; 
int num = 1,head[maxn],dfn[maxn] ; 
inline void add_edge(int u,int v,int w) { 
     edge[++ num].v = v;edge[num].next = head[u];head[u] = num; edge[num].w = w;} 
int h[maxn]; 
int num1 = 0; 
inline void add(int u,int v) { 
    if(u == v) return ; 
    e[++ num1].v = v;e[num1].next = h[u];h[u] = num1; 
} 
int n,m,id = 0,dad[maxn][20],deep[maxn];LL mn[maxn];    
void dfs(int x) {
    dfn[x] = ++id; 
    deep[x] = deep[dad[x][0]] + 1;
    for(int i = 0;dad[x][i];++ i) dad[x][i + 1] = dad[dad[x][i]][i]; 
    for(int i = head[x];i;i = edge[i].next) {
        int v = edge[i].v;
        if(v == dad[x][0]) continue; 
        dad[v][0] = x; 
        mn[v] = std::min(mn[x],(LL)edge[i].w); 
        dfs(v); 
    } 
} 
int lca(int x,int y) { 
    if(deep[x] > deep[y]) std::swap(x,y) ; 
    for(int i = 18;i >= 0;-- i) if(deep[dad[y][i]] >= deep[x]) y = dad[y][i];
        if(x == y) return x; 
    for(int i = 18;i >= 0;-- i) 
        if(dad[x][i] != dad[y][i]) x = dad[x][i],y = dad[y][i]; 
    return dad[x][0]; 
} 
int tt[maxn],stack[maxn];  
inline bool cmp(int x,int y) {  return dfn[x] < dfn[y]; } 
LL dp[maxn]; 
void Dp(int x) {
    dp[x] = (LL)mn[x]; 
    LL tmp = 0; 
    for(int i = h[x];i;i = e[i].next) { 
        Dp(e[i].v);
        tmp += dp[e[i].v]; 
    } 
    h[x] = 0;//顺便清空
    if(tmp == 0) dp[x] = (LL)mn[x]; 
    else dp[x] = std::min(dp[x],tmp);//要么割自己，要么割子树 
} 
void solve(int k = 0,int top = 0,int tot = 0) {
    k = read(); 
    for(int i = 1;i <= k;++ i) tt[i] = read(); 
    std::sort(tt + 1,tt + k + 1,cmp); 
    stack[++ tot] = tt[1]; 
    for(int i = 2;i <= k;++ i) 
        if(lca(tt[tot],tt[i]) != tt[tot]) tt[++ tot] = tt[i]; 
    //根据题意，切掉子树的根后子树就不用管了 
    stack[++ top] = 1;  
    for(int i = 1;i <= tot;++ i) {  
        int now = tt[i],f = lca(now,stack[top]);   
        while("tle") { 
            if(deep[stack[top - 1]] <= deep[f]) {  
                add(f,stack[top --]);  
                    if(stack[top] != f)stack[++ top] = f; 
                break;   
            }  
            add(stack[top - 1],stack[top]) ;top --;  
        } 
        if(stack[top] != now)stack[++ top] = now;  
    }  
    while(-- top) add(stack[top],stack[top + 1]);  
    Dp(1);  
    printf("%lld
",dp[1]);      
} 
int main() { 
    n = read();  
    for(int u,v,w,i = 1;i < n;++ i) { 
        u = read();v = read();w = read(); 
        add_edge(u,v,w); add_edge(v,u,w); 
    } 
    mn[1] = 10000000000000007LL; 
    dfs(1);
    m = read(); 
    for(int i = 1;i <= m;++ i) { 
        solve();
    } 
    return 0;   
}