ZOJ 4097 Rescue the Princess

在这个物欲横流的社会

oj冷漠无情

只有这xx还有些温度

越界就越界吧  wrong 怎么回事。。。。

给出一个图

然后给出q次询问

问是否存在v和w分别到u的路径且边不重复

在边双连通分量中 任意两点 都至少存在两条边不重复的路径

那么若u  v w 在同一连通图中

则存在

若不在

则只有u在 v 和 w 中间时才存在

想一想是不是

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define rb(a) scanf("%lf", &a)
#define rf(a) scanf("%f", &a)
#define pd(a) printf("%d\n", a)
#define plld(a) printf("%lld\n", a)
#define pc(a) printf("%c\n", a)
#define ps(a) printf("%s\n", a)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 101000, INF = 0x7fffffff, maxm = 1000100, maxh = 20;
int n, m, q;
int head[maxn], nex[maxm], cnt, head2[maxn], cnt2;
int pre[maxn], low[maxn], sccno[maxn], dfs_clock, scc_cnt;
int pa[maxn], anc[maxn][maxh + 1], deep[maxn], instack[maxn];
int f[maxn];
stack<int> s;

struct node
{
    int u, v;
}Node[maxm];

void add_(int u, int v)
{
    Node[cnt].u = u;
    Node[cnt].v = v;
    nex[cnt] = head[u];
    head[u] = cnt++;
}

void add(int u, int v)
{
    add_(u, v);
    add_(v, u);
}

struct edge{
    int u, v, next;
}Edge[maxm];

void add2(int u, int v)
{
    Edge[cnt2].u = u;
    Edge[cnt2].v = v;
    Edge[cnt2].next = head2[u];
    head2[u] = cnt2++;
}

void tarjan(int u,int fa){
    low[u] = pre[u] = ++dfs_clock;
    s.push(u);
    instack[u] = 1;
    for(int i = head[u]; i != -1; i = nex[i])
    {
        int v = Node[i].v;
        if(i == (fa ^ 1)) continue;
        if(!pre[v])
        {
            tarjan(v, i);
            low[u] = min(low[u], low[v]);
        }
        else if(instack[v])
            low[u] = min(low[u], pre[v]);

    }
    if(pre[u] == low[u])
    {
        scc_cnt++;
        for(;;)
        {
           int x = s.top(); s.pop();
           sccno[x] = scc_cnt;
           instack[x] = 0;
           if(x == u) break;
        }
    }
}

int dfs(int u, int fa)
{
    for(int i = 1; i < maxh; i++)
        anc[u][i] = anc[anc[u][i - 1]][i - 1];
    for(int i = head2[u]; i != -1; i = Edge[i].next)
    {
        int v = Edge[i].v;
        if(v == fa || deep[v]) continue;
        anc[v][0] = u;
        deep[v] = deep[u] + 1;
        dfs(v, u);
    }
}

int lca(int u, int v)
{
    if(deep[u] < deep[v]) swap(u, v);
    for(int i = maxh; i >= 0; i--)
        if(deep[anc[u][i]] >= deep[v])
            u = anc[u][i];

    for(int i = maxh; i >= 0; i--)
    {
        if(anc[u][i] != anc[v][i])
        {
            u = anc[u][i];
            v = anc[v][i];
        }
    }
    if(u == v) return u;
    return anc[u][0];
}

int find(int x)
{
    return f[x] == x ? x : (f[x] = find(f[x]));
}

void init()
{
    rep(i, 1, maxn) f[i] = i;
    mem(head, -1);
    mem(head2, -1);
    mem(deep, 0);
    dfs_clock = scc_cnt = cnt = cnt2 = 0;
    mem(sccno, 0); mem(pre, 0);
    mem(low, 0);
    mem(anc, 0);
    mem(instack, 0);
}

int main()
{
    int T;
    rd(T);
    while(T--)
    {
        init();
        rd(n), rd(m), rd(q);
        rap(i, 1, m)
        {
            int u, v;
            rd(u), rd(v);
            if(u == v) continue;
            add(u, v);
        }
        for(int i = 1; i <= n; i++) if(!pre[i]) tarjan(i, -1);
        rap(u, 1, n)
        {
            for(int i = head[u]; i != -1; i = nex[i])
            {
                int v = Node[i].v;
                if(sccno[u] != sccno[v])
                {
                    add2(sccno[u], sccno[v]);
                    int l = find(sccno[u]), r = find(sccno[v]);
                    if(l == r) continue;
                    f[l] = r;
                }
            }
        }
        rap(i, 1, scc_cnt)
            if(!deep[i]) deep[i] = 1, dfs(i, -1);
        rap(i, 1, q)
        {
            int u, v, w;
            rd(u), rd(v), rd(w);
            if(find(sccno[u]) != find(sccno[v]) || find(sccno[u]) != find(sccno[w]))
            {
                puts("No");
                continue;
            }
            if(sccno[v] != sccno[w])
            {
                int lc = lca(sccno[v], sccno[w]);
                int ans1 = lca(sccno[u], sccno[v]);
                int ans2 = lca(sccno[u], sccno[w]);
                int ans3 =lca(lc, sccno[u]);
                if(ans3 == lc && (ans1 == sccno[u] || ans2 == sccno[u]))
                {
                    puts("Yes");
                }
                else
                {
                    puts("No");
                }
            }
            else
            {
                if(sccno[u] == sccno[v] )
                {
                    puts("Yes");
                    continue;
                }
                else
                    puts("No");
            }
        }
    }

    return 0;
}

原文地址：https://www.cnblogs.com/WTSRUVF/p/10727360.html