HDU 5409 CRB and Graph 双连通缩点 + st表

HDU 5409

显然要先双连通缩成一棵树。 然后对于树上的边才有答案。

对于一条边来说， 两侧的最大值分为mx1 , mx2 ， 那么 u 一定是min(mx1, mx2)， v 一定是 u + 1。

这个经过仔细分析一下就能得到， 然后按dfs序建个st表就好啦。

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 1e5 + 2;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;}

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int n, m;
vector<PII> G[N];
vector<PII> bG[N];

int stk[N], top;
int dfn[N], low[N], belong[N], idx, cnt;
int mx[N];
bool brige[N];
PII ans[N];
int in[N], ot[N];

int Log[N];
struct ST {
    int dp[N][20]; int ty;
    void build(int n, int b[], int _ty) {
        ty = _ty;
        for(int i = -(Log[0]=-1); i < N; i++)
        Log[i] = Log[i - 1] + ((i & (i - 1)) == 0);
        for(int i = 1; i <= n; i++) {
            dp[in[i]][0] = mx[i];
        }
        for(int j = 1; j <= Log[n]; j++)
            for(int i = 1; i+(1<<j)-1 <= n; i++)
                dp[i][j] = max(dp[i][j-1], dp[i+(1<<(j-1))][j-1]);
    }
    inline int query(int x, int y) {
        if(x > y) return 0;
        int k = Log[y - x + 1];
        return ty * max(dp[x][k], dp[y-(1<<k)+1][k]);
    }
} rmq;

void tarjan(int u, int pid) {
    dfn[u] = low[u] = ++idx;
    stk[++top] = u;
    for(auto &e : G[u]) {
        if(e.fi == pid) continue;
        if(!dfn[e.se]) {
            tarjan(e.se, e.fi);
            chkmin(low[u], low[e.se]);
            if(dfn[u] < low[e.se]) {
                brige[e.fi] = true;
            }
        } else {
            chkmin(low[u], dfn[e.se]);
        }
    }
    if(dfn[u] == low[u]) {
        cnt++;
        while(1) {
            int v = stk[top--];
            belong[v] = cnt;
            chkmax(mx[cnt], v);
            if(v == u) break;
        }
    }
}

void go(int u, int fa) {
    in[u] = ++idx;
    for(auto &e : bG[u]) {
        if(e.se == fa) continue;
        go(e.se, u);
    }
    ot[u] = idx;
}

void solve(int u, int fa) {
    for(auto &e : bG[u]) {
        int v = e.se, eid = e.fi;
        if(v == fa) continue;
        int mx1 = rmq.query(in[v], ot[v]);
        int mx2 = max(rmq.query(1, in[v] - 1), rmq.query(ot[v] + 1, cnt));
        ans[eid].fi = min(mx1, mx2);
        ans[eid].se = ans[eid].fi + 1;
        solve(v, u);
    }
}

int main() {

    int T; scanf("%d", &T);
    while(T--) {
        scanf("%d%d", &n, &m);
        idx = cnt = top = 0;
        for(int i = 1; i <= n; i++) {
            G[i].clear();
            bG[i].clear();
            dfn[i] = low[i] = 0;
            mx[i] = 0;
        }
        for(int i = 1; i <= m; i++) {
            int u, v;
            scanf("%d%d", &u, &v);
            G[u].push_back(mk(i, v));
            G[v].push_back(mk(i, u));
            brige[i] = false;
            ans[i] = mk(0, 0);
        }

        for(int i = 1; i <= n; i++) {
            if(!dfn[i]) {
                tarjan(i, 0);
            }
        }
        for(int u = 1; u <= n; u++) {
            for(auto &e : G[u]) {
                if(belong[u] == belong[e.se]) continue;
                bG[belong[u]].push_back(mk(e.fi, belong[e.se]));
            }
        }

        idx = 0;
        go(1, 0);

        rmq.build(cnt, mx, 1);

        solve(1, 0);

        for(int i = 1; i <= m; i++) {
            printf("%d %d
", ans[i].fi, ans[i].se);
        }
    }
    return 0;
}

/*
*/