洛谷 [P3496] BLO

割点

首先 tarjan 求割点,
对于不是割点的点, 答案是 2 * (n-1) 有序，所以要乘 2
对于是割点的点, 答案是删去该点后所有连通块的个数加上 n-1 在乘 2

#include 
#include 
#include 
#include 
#include 
#define ll long long
using namespace std;
const int MAXN = 1000005;
ll ans[MAXN];
int head[MAXN], n, m, nume, dfn[MAXN], low[MAXN], ind, siz[MAXN];
bool f[MAXN];
struct edge{
    int to, nxt;
}e[MAXN];
void adde(int from, int to) {
    e[++nume].to = to;
    e[nume].nxt = head[from];
    head[from] = nume;
}
int init() {
    int rv = 0, fh = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') fh = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        rv = (rv<<1) + (rv<<3) + c - '0';
        c = getchar();
    }
    return fh * rv;
}
void tarjan(int u) {
    dfn[u] = low[u] = ++ind;siz[u] = 1;
    int flag = 0, sum = 0;
    for(int i = head[u]; i; i = e[i].nxt) {
        int v = e[i].to;
        if(!dfn[v]) {
            tarjan(v);
            siz[u] += siz[v];
            low[u] = min(low[u], low[v]);
            if(low[v] >= dfn[u]) {
                flag++;
                ans[u] += (ll) siz[v] * (n - siz[v]);
                sum += siz[v];
                if(u != 1 || flag > 1) {
                    f[u] = 1;
                }
            }
        }else low[u] = min(low[u], dfn[v]);
    }
    if(f[u]) ans[u] += (ll) (n - sum - 1) * (sum + 1) + n - 1;
    else ans[u] = 2 * n - 2;
}
int main() {
    n = init(); m = init();
    for(int i = 1; i <= m; i++) {
        int u = init(), v = init();
        if(u == v) continue;
        adde(u, v); adde(v, u);
    }
    tarjan(1);
    for(int i = 1; i <= n; i++) printf("%lld\n", ans[i]);
    return 0;
}

转载于:https://www.cnblogs.com/Mr-WolframsMgcBox/p/9095086.html