Day10 - A - Rescue the Princess ZOJ - 4097

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 $l_ili-th island and the r_iri-th island. The knights can go through each bridge in both directions.$

Landing separately on the

Thus, to successfully bring back the princess, two paths \textbf{with no common bridges} are needed: one starts from the

As the princess is caught very often, the knights will ask you for help

Input

There are multiple test cases. The first line of the input contains an integer

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

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

The following $u_iui, v_ivi and w_iwi (1 \le u_i,v_i,w_i \le n1≤ui,vi,wi≤n), indicating the island where the princess is imprisoned and the starting islands of the two knights.$

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

Output

For each test case output

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+5;

int dfn[maxm], low[maxm], s[maxm], dfscnt, top, bcc, bccnum[maxm], block[maxm], father[maxm], blocknum, depth[maxm], grand[maxm][25];
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()-1);
    edges.push_back(Node{v, u});
    G2[v].push_back(edges.size()-1);
}

void init(int n) {
    for(int i = 0; i <= n; ++i) G[i].clear(), G2[i].clear();
    edges.clear();
    dfscnt = top = 0;
    bcc = blocknum = 1;
    memset(vis, 0, sizeof(vis));
    memset(block, 0, sizeof(block)), memset(father, 0, sizeof(father));
    memset(dfn, 0, sizeof(dfn)), memset(low, 0, sizeof(low));
    memset(bccnum, 0, sizeof(bccnum));
}

void tarjan(int u, int fa) {    //边双缩点
    int v, k = 0, siz = G[u].size();
    dfn[u] = low[u] = ++dfscnt;
    s[++top] = u;
    father[u] = fa;
    for(int i = 0; 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 = 0; i < siz; ++i) {
        int v = G[u][i];
        if(!block[v])
            dfs1(v);
    }
}

void dfs2(int u, int fa) {   // build lca tree
    vis[u] = 1;
    depth[u] = depth[fa] + 1;
    grand[u][0] = fa;
    for(int i = 1; i <= N; ++i)
        grand[u][i] = grand[grand[u][i-1]][i-1];
    int siz = G2[u].size();
    for(int i = 0; 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 >= 0; --i)
        if(depth[a] <= depth[b] - (1<<i)) b = grand[b][i];
    if(a == b) return a;
    for(int i = N; i >= 0; --i) {
        if(grand[a][i] == grand[b][i]) continue;
        else {
            a = grand[a][i], b = grand[b][i];
        }
    }
    return grand[a][0];
}

void run_case() {
    int n, m, q, u, v, w;
    cin >> n >> m >> q;
    init(n);
    for(int i = 0; i < m; ++i) {
        cin >> u >> v;
        G[u].push_back(v), G[v].push_back(u);
    }
    for(int i = 1; i <= n; ++i)  //找连通块
        if(!block[i]) {
            dfs1(i);
            blocknum++;
        }
    for(int i = 1; i <= n; ++i) // 缩点
        if(!dfn[i])
            tarjan(i, i);
    N = floor(log(bcc + 0.0) / log(2.0)) + 1; //最多能跳的2^i祖先
    for(int i = 1; i <= n; ++i) {
        int v = father[i];
        if(bccnum[i] != bccnum[v])
            addedge(bccnum[i], bccnum[v]);
    }
    // build lca tree
    for(int i = 1; i < bcc; ++i) {
        if(!vis[i]) {
            dfs2(i, 0);
        }
    }
    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(0);
    int t;
    cin >> t;
    while(t--)
        run_case();
    return 0;
}

View Code

(自用板)

posted @ 2020-02-02 14:23 GRedComeT 阅读(140) 评论(0) 编辑收藏举报

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