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传送门
分析
我们先去把每一个相同颜色的联通快记录下来,然后去遍历每一个黑点,计算他连接的所有白点所在的联通块的大小,有两种情况
- 黑点就是端点
这种情况只需要答案加上白色联通快大小的和即可 - 两个端点都是白点
任意两个联通快组合即可
代码
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const int INF = 0x3f3f3f3f;
const int N = 2e6 + 10;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
int h[N],e[N],ne[N],idx;
int col[N];
char str[N];
int id[N];
ll num[N];
bool st[N];
int tot;
int n;void add(int x,int y){ne[idx] = h[x],e[idx] = y,h[x] = idx++;
}void dfs(int u,int c,int fa){if(col[u] != c) return;num[tot]++;st[u] = true;id[u] = tot;for(int i = h[u];~i;i = ne[i]){int j = e[i];if(j == fa) continue;dfs(j,c,u);}
}int main() {memset(h,-1,sizeof h);read(n);scanf("%s",str + 1);for(int i = 1;i < n;i++){int x,y;read(x),read(y);add(x,y),add(y,x);}for(int i = 1;i <= n;i++) if(str[i] == 'B') col[i] = 1;for(int i = 1;i <= n;i++){if(st[i]) continue;tot++;dfs(i,col[i],-1);}ll res = 0;for(int i = 1;i <= n;i++){if(!col[i]) continue;ll sum = 0,p = 0;for(int j = h[i];~j;j = ne[j]){int v = e[j];if(col[v]) continue;sum += num[id[v]];}for(int j = h[i];~j;j = ne[j]){int v = e[j];if(col[v]) continue;p += num[id[v]] * (sum - num[id[v]]);}res += sum;p /= 2;res += p;}dl(res);return 0;
}/**7
BBWWWWW
1 2
1 3
3 4
2 5
5 6
2 7*/
/**
* ┏┓ ┏┓+ +
* ┏┛┻━━━┛┻┓ + +
* ┃ ┃
* ┃ ━ ┃ ++ + + +
* ████━████+
* ◥██◤ ◥██◤ +
* ┃ ┻ ┃
* ┃ ┃ + +
* ┗━┓ ┏━┛
* ┃ ┃ + + + +Code is far away from
* ┃ ┃ + bug with the animal protecting
* ┃ ┗━━━┓ 神兽保佑,代码无bug
* ┃ ┣┓
* ┃ ┏┛
* ┗┓┓┏━┳┓┏┛ + + + +
* ┃┫┫ ┃┫┫
* ┗┻┛ ┗┻┛+ + + +
*/
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