HDU - 5354 Bipartite Graph 分治 + 并查集 （看题解）

HDU - 5354

如果询问一个点的话， 直接用并查集check一下就好了。

要求n个点都求出答案的话， 用分治优化一下， 感觉挺巧妙的。

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#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 + 7;
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<int> G[N];
int ans[N];
int fa[N], sz[N], xr[N];
int top;
PII stk[N];
int txr[N];

PII getRoot(int x) {
    if(fa[x] == x) return mk(0, x);
    PII ret = getRoot(fa[x]);
    ret.fi ^= xr[x];
    return ret;
}

bool unite(int u, int v) {
    PII x = getRoot(u);
    PII y = getRoot(v);
    if(x.se == y.se) {
        return (x.fi ^ y.fi);
    }
    else {
        if(sz[x.se] < sz[y.se]) swap(x, y);
        fa[y.se] = x.se;
        xr[y.se] = (x.fi ^ y.fi ^ 1);
        sz[x.se] += sz[y.se];
        stk[++top] = mk(x.se, y.se);
        txr[top] = xr[y.se];
        assert(top < N);
        return true;
    }
}

inline void rollback() {
    int x = stk[top].fi;
    int y = stk[top].se;
    xr[y] = txr[top];
    top--;
    sz[x] -= sz[y];
    fa[y] = y;
}


void init() {
    top = 0;
    for(int i = 1; i <= n; i++) {
        G[i].clear();
        fa[i] = i;
        sz[i] = 1;
        xr[i] = 0;
    }
}

void solve(int l, int r) {
    if(l == r) {
        ans[l] = 1;
        return;
    }
    int preTop = top;
    int mid = l + r >> 1;
    bool can = true;

    for(int u = mid + 1; u <= r; u++) {
        for(auto &v : G[u]) {
            if(v >= l && v <= mid) continue;
            if(!unite(u, v)) {
                can = false;
                u = r + 1;
                break;
            }
        }
    }

    if(!can) {
        for(int u = l; u <= mid; u++) {
            ans[u] = 0;
        }
    }
    else {
        solve(l, mid);
    }

    while(top > preTop) rollback();

    can = true;

    for(int u = l; u <= mid; u++) {
        for(auto &v : G[u]) {
            if(v > mid && v <= r) continue;
            if(!unite(u, v)) {
                can = false;
                u = r + 1;
                break;
            }
        }
    }

    if(!can) {
        for(int u = mid + 1; u <= r; u++) {
            ans[u] = 0;
        }
    }
    else {
        solve(mid + 1, r);
    }

    while(top > preTop) rollback();
}

int main() {

    int T; scanf("%d", &T);
    while(T--) {
        scanf("%d%d", &n, &m);
        init();
        for(int i = 1; i <= m; i++) {
            int u, v;
            scanf("%d%d", &u, &v);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        solve(1, n);

        for(int i = 1; i <= n; i++) {
            if(ans[i]) putchar('1');
            else putchar('0');
        }
        puts("");
    }
    return 0;
}

/*
*/