bzoj4543[POI2014]Hotel
题目链接
题解
这不是裸地点分嘛 ,我真傻,真的
n^2 这不是是sb题,~滑稽 ~
枚举点转换为无根树,暴力子树中点的深度
计数转移
令a b c d为已知四颗子树,则新来一颗深度为k的点数为e的子树
推下式子
\(new_ans=e*(a*b+a*c+b*c+d*a+d*b+d*c)\)
用s维护 \((a*b+a*c+b*c+d*a+d*b+d*c .....)\)
用cnt维护\((a+b+c+d+e+....)\)
s的转移为 \(s + cnt * tmp\);
代码
#include<algorithm>
#include<cstring>
#include<cstdio>
inline int read() {
int x = 0,f = 1;
char c = getchar();
while(c < '0' || c >'9') {if(c == '-') f = -1;c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0',c = getchar();
return x * f;
}
int n;
const int maxn = 5007;
struct Node {
int v,next;
Node(int a = 0,int b = 0): v(a),next(b) {};
}edge[maxn << 1];
int num,head[maxn];
inline void add_edge(int u,int v) {
edge[++num] = Node(v,head[u]);head[u] = num;
}
int dp[maxn][maxn];
int deep[maxn],mx = 0,cnt[maxn],tmp[maxn],s[maxn];
void dfs(int x,int fa) {
tmp[deep[x]] ++;
mx = std::max(mx,deep[x]);
for(int i = head[x];i;i = edge[i].next) {
int v = edge[i].v;
if(v == fa) continue;
deep[v] = deep[x] + 1;
dfs(v,x);
}
}
int main() {
n = read();
for(int u,v,i = 1;i < n;++ i) {
u = read() , v = read();
add_edge(u,v);add_edge(v,u);
}
long long int ans = 0;
for(int i = 1;i <= n;++ i) {
for(int j = head[i];j;j = edge[j].next) {
int v = edge[j].v;
mx = 0;
deep[v] = 1;
dfs(v,i);
for(int k = 1;k <= mx;++ k) {
ans += s[k] * tmp[k];
s[k] += tmp[k] * cnt[k];
cnt[k] += tmp[k];
tmp[k] = 0;
}
}
memset(s,0,sizeof s);
memset(cnt,0,sizeof cnt);
}
printf("%lld\n",ans);
return 0;
}