BZOJ 3331 [BeiJing2013]压力-Tarjan + 树上差分

Solution

Tarjan 点双缩点， 加上树上差分计算。

注意特判。。。 我特判挂了好久呜呜呜

Code

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#define rd read()
using namespace std;

const int N = 1e5 + 5;
const int M = 2e5 + 5;
const int base = 30;

int head[N], tot;
int Head[N << 1], Tot;
int low[N], dfn[N], cnt, col, c[N], cut[N], n, m, Q;
int f[N << 2][30], nd, id[N], idf[N << 2], dep[N << 2];
int st[N], tp, num[N], sum[N << 2];

vector<int> q[N << 1];

struct edge {
    int nxt, to;
}e[M << 1], E[M << 3];

int read() {
    int X = 0, p = 1; char c = getchar();
    for (; c > '9' || c < '0'; c = getchar()) if (c == '-') p = -1;
    for (; c >= '0' && c <= '9'; c = getchar()) X = X * 10 + c - '0';
    return X * p;
}

void add(int u, int v) {
    e[++tot].to = v;
    e[tot].nxt = head[u];
    head[u] = tot;
}

void Add(int u, int v) {
    E[++Tot].to = v;
    E[Tot].nxt = Head[u];
    Head[u] = Tot;
}

void tarjan(int u) {
    low[u] = dfn[u] = ++cnt;
    st[++tp] = u;
    int flag = 0;
    for (int i = head[u]; i; i = e[i].nxt) {
        int nt = e[i].to;
        if (!dfn[nt]) {
            tarjan(nt);
            low[u] = min(low[u], low[nt]);
            if (low[nt] >= dfn[u]) {
                flag++;
                if (flag > 1 || u - 1)
                    cut[u] = 1;
                col++;
                for (; tp;) {
                    int v = st[tp--];
                    q[col].push_back(v);
                    if (v == nt)
                        break;
                }
                q[col].push_back(u);
            }
        }
        else low[u] = min(low[u], dfn[nt]);
    }
}

void dfs(int u) {
    for (int i = Head[u]; i; i = E[i].nxt) {
        int nt = E[i].to;
        if (f[u][0] == nt)
            continue;
        f[nt][0] = u;
        dep[nt] = dep[u] + 1;
        dfs(nt);
    }
}

int LCA(int x, int y) {
    if (dep[x] < dep[y])
        swap(x, y);
    for (int k = base - 1; ~k; --k)
        if (dep[f[x][k]] >= dep[y])
            x = f[x][k];
    if (x == y)
        return x;
    for (int k = base - 1; ~k; --k) 
        if (f[x][k] != f[y][k])
            x = f[x][k], y = f[y][k];
    return f[x][0];
}

void dfs2(int u) {
    for (int i = Head[u]; i; i = E[i].nxt) {
        int nt = E[i].to;
        if (nt == f[u][0]) 
            continue;
        dfs2(nt);
        sum[u] += sum[nt];
    }
    if(u > col)
        num[idf[u]] += sum[u];
}

int main()
{
    //freopen("1.in","r", stdin);
    //freopen("out.out","w",stdout);
    n = rd; m = rd; Q = rd;
    for (int i = 1; i <= m; ++i) {
        int u = rd, v = rd;
        add(u, v); add(v, u);
    }
    tarjan(1);
    nd = col;
    for (int i = 1; i <= n; ++i)
        if (cut[i]) c[i] = ++nd, idf[nd] = i;
    for (int i = 1; i <= col; ++i) 
        for(int j = 0, len = q[i].size(); j < len; ++j) {
            int x = q[i][j];
            if (cut[x]) 
                Add(i, c[x]), Add(c[x], i);
            else c[x] = i;
        }
    dep[1] = 1;
    dfs(1);
    for (int k = 1; k < base; ++k)
        for (int i = 1; i <= nd; ++i)
            f[i][k] = f[f[i][k - 1]][k - 1];
    for (; Q; Q--) {
        int u = rd, v = rd, lca;
        if (c[u] <= col) num[u]++; 
        if (c[v] <= col) num[v]++;
        u = c[u]; v = c[v];
        if (u == v) continue;
        lca = LCA(u, v);
        sum[v]++; sum[u]++;
        sum[lca]--; sum[f[lca][0]]--;
    }
    dfs2(1);
    for (int i = 1; i <= n; ++i)
        printf("%d
", num[i]);
}