Day10 - A - Rescue the Princess ZOJ

Princess Cjb is caught by Heltion again! Her knights Little Sub and Little Potato are going to Heltion Kingdom to rescue her.

Heltion Kingdom is composed of nn islands, numbered from 11 to nn. There are mm bridges in the kingdom, among which the ii-th bridge connects the l_ili-th island and the r_iri-th island. The knights can go through each bridge in both directions.

Landing separately on the vv-th and the ww-th island, the two knights start their journey heading to the uu-th island where the princess is imprisoned. However, as the knights are fat and the bridges are unstable, there will be a risk of breaking down the bridge and falling into the water if they go through one or more common bridges during their journey.

Thus, to successfully bring back the princess, two paths \textbf{with no common bridges} are needed: one starts from the vv-th island and leads to the uu-th island, while the other starts from the ww-th island and also leads to the uu-th island.

As the princess is caught very often, the knights will ask you for help qq times. Each time, given their starting islands and their goal, you need to tell them whether it's possible to find two paths satisfying the constraints above.

Input

There are multiple test cases. The first line of the input contains an integer TT, indicating the number of test cases. For each test case:

The first line contains three integers nn, mm and qq (1 \le n \le 10^51≤n≤105, 0 \le m \le 2 \times 10^50≤m≤2×105, 1 \le q \le 10^51≤q≤105), indicating the number of islands, the number of bridges and the number of queries.

The following mm lines describe the bridges. The ii-th line contains two integers l_ili and r_iri (1 \le l_i,r_i \le n1≤li,ri≤n), indicating the two islands the ii-th bridge connects. Notice that different bridges may connect the same pair of islands and a bridge may connect an island to itself.

The following qq lines describe the queries. The ii-th line contains three integers u_iui, v_ivi and w_iwi (1 \le u_i,v_i,w_i \le n1≤ui,vi,wi≤n), indicating the island where the princess is imprisoned and the starting islands of the two knights.

It's guaranteed that the sum of nn of all test cases will not exceed 5 \times 10^55×105, the sum of mm of all test cases will not exceed 10^6106, and the sum of qq of all test cases will not exceed 5 \times 10^55×105.

Output

For each test case output qq lines indicating the answers of the queries. For each query, if two paths meeting the constraints can be found, output "Yes" (without quotes), otherwise output "No" (without quotes).

Sample Input

Sample Output

No

Yes

Yes

Yes

Yes

Yes

Hint

For the first sample test case:

For the 2nd query, we can select the paths 4-1 and 2-1.
For the 3rd query, we can select the paths 2-1 and 3-1.
For the 4th query, we can select the paths 3-1 and 3-2-1.

For the second sample test case:

For the 1st query, as the knights and the princess are on the same island initially, the answer is "Yes".
For the 2nd query, as one of the knights are on the same island with the princess initially, he does not need to cross any bridge. The other knight can go from island 1 to island 2 directly.

思路：求两个点不同边路径，想到边双缩点，这题需要考虑重边和多个连通块，对每个连通块缩点，跑lca，判断lca即可

#include<iostream>

#include<cstdio>

#include<cstring>

#include<queue>

#include<algorithm>

#include<cmath>

#include<set>

#include<vector>

#include<map>

using namespace std;

#define lowbit(x) ((x)&(-x))

typedef long long LL;

typedef pair<LL, LL> PLL;

const int maxm = 2e5+;

int dfn[maxm], low[maxm], s[maxm], dfscnt, top, bcc, bccnum[maxm], block[maxm], father[maxm], blocknum, depth[maxm], grand[maxm][];

int vis[maxm], N;

struct Node {

    int u, v;

};

vector<int> G[maxm], G2[maxm];

vector<Node> edges;

void addedge(int u, int v) {

    edges.push_back(Node{u, v});

    G2[u].push_back(edges.size()-);

    edges.push_back(Node{v, u});

    G2[v].push_back(edges.size()-);

}

void init(int n) {

    for(int i = ; i <= n; ++i) G[i].clear(), G2[i].clear();

    edges.clear();

    dfscnt = top = ;

    bcc = blocknum = ;

    memset(vis, , sizeof(vis));

    memset(block, , sizeof(block)), memset(father, , sizeof(father));

    memset(dfn, , sizeof(dfn)), memset(low, , sizeof(low));

    memset(bccnum, , sizeof(bccnum));

}

void tarjan(int u, int fa) {    //边双缩点

    int v, k = , siz = G[u].size();

    dfn[u] = low[u] = ++dfscnt;

    s[++top] = u;

    father[u] = fa;

    for(int i = ; i < siz; ++i) {

        v = G[u][i];

        if(v == fa && !k) {  // 判重边

            k++;

            continue;

        }

        if(!dfn[v]) {

            tarjan(v, u);

            low[u] = min(low[u], low[v]);

        } else

            low[u] = min(low[u], dfn[v]);

    }

    if(dfn[u] == low[u]) {

        do {

            v = s[top--];

            bccnum[v] = bcc;

        } while(u != v);

        bcc++;

    }

}

void dfs1(int u) {

    block[u] = blocknum;

    int siz = G[u].size();

    for(int i = ; i < siz; ++i) {

        int v = G[u][i];

        if(!block[v])

            dfs1(v);

    }

}

void dfs2(int u, int fa) {   // build lca tree

    vis[u] = ;

    depth[u] = depth[fa] + ;

    grand[u][] = fa;

    for(int i = ; i <= N; ++i)

        grand[u][i] = grand[grand[u][i-]][i-];

    int siz = G2[u].size();

    for(int i = ; i < siz; ++i) {

        int v = edges[G2[u][i]].v;

        if(v != fa) {

            dfs2(v, u);

        }

    }

}

int lca(int a, int b) {

    if(a == b) return a;

    if(depth[a] > depth[b]) swap(a, b);

    for(int i = N; i >= ; --i)

        if(depth[a] <= depth[b] - (<<i)) b = grand[b][i];

    if(a == b) return a;

    for(int i = N; i >= ; --i) {

        if(grand[a][i] == grand[b][i]) continue;

        else {

            a = grand[a][i], b = grand[b][i];

        }

    }

    return grand[a][];

}

void run_case() {

    int n, m, q, u, v, w;

    cin >> n >> m >> q;

    init(n);

    for(int i = ; i < m; ++i) {

        cin >> u >> v;

        G[u].push_back(v), G[v].push_back(u);

    }

    for(int i = ; i <= n; ++i)  //找连通块

        if(!block[i]) {

            dfs1(i);

            blocknum++;

        }

    for(int i = ; i <= n; ++i) // 缩点

        if(!dfn[i])

            tarjan(i, i);

    N = floor(log(bcc + 0.0) / log(2.0)) + ; //最多能跳的2^i祖先

    for(int i = ; i <= n; ++i) {

        int v = father[i];

        if(bccnum[i] != bccnum[v])

            addedge(bccnum[i], bccnum[v]);

    }

    // build lca tree

    for(int i = ; i < bcc; ++i) {

        if(!vis[i]) {

            dfs2(i, );

        }

    }

    while(q--) {

        cin >> u >> v >> w;

        // 不同连通块

        if(block[u] != block[v] || block[u] != block[w]) {

            cout << "No" << "\n";

            continue;

        }

        u = bccnum[u], v = bccnum[v], w = bccnum[w];

        // 在同一连通块 不同点

        if(u == v || u == w) {

            cout << "Yes" << "\n";

            continue;

        }

        if(v == w) {

            cout << "No" << "\n";

            continue;

        }

        int uv, uw, vw, uvw;

        // u是vw的根即可

        uv = lca(u, v), uw = lca(u, w), vw = lca(v, w), uvw = lca(uw, v);

        if(vw == uvw && (uv == u || uw == u))

            cout << "Yes" << "\n";

        else

            cout << "No" << "\n";

    }

}

int main() {

    ios::sync_with_stdio(false), cin.tie();

    int t;

    cin >> t;

    while(t--)

        run_case();

    return ;

}

(自用板)