hdu 3686 Traffic Real Time Query System 点双两通分量 + LCA。这题有重边！！！

http://acm.hdu.edu.cn/showproblem.php?pid=3686

我要把这题记录下来。

一直wa。

自己生成数据都是AC的。现在还是wa。留坑。

我感觉我现在倒下去床上就能睡着了。

不知道是我的LCA错了，还是tarjan

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;


#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
const int maxm = 200000 + 20;
const int maxn = 10000 + 20;
struct Edge {
    int u, v, id, tonext;
} e[maxm * 2];
int first[maxm], num;
void addEdge(int u, int v, int id) {
    ++num;
    e[num].u = u, e[num].v = v, e[num].tonext = first[u];
    first[u] = num;
    e[num].id = id;
}
int uuu[maxm], vvv[maxm];
int n, m;
int low[maxm], DFN[maxm], st[maxm], id[maxm];
int top, when, toSelid;
bool iscut[maxm];
vector<int>bolg[maxm];
int root;
void tarjan(int cur, int fa) {
    DFN[cur] = low[cur] = ++when;
    int child = 0;
    for (int i = first[cur]; i; i = e[i].tonext) {
        int v = e[i].v;
        if (v == fa) continue;
        if (!DFN[v]) {
            child++;
            st[++top] = e[i].id;
            tarjan(v, cur);
            low[cur] = min(low[cur], low[v]);
            if (low[v] >= DFN[cur]) {
                iscut[cur] = true;
//                if (cur == root && child < 2) iscut[cur] = false;
                ++toSelid;
                do {
                    int eID = st[top--];
                    bolg[uuu[eID]].push_back(toSelid);
                    bolg[vvv[eID]].push_back(toSelid);
                    id[eID] = toSelid;
                } while (st[top + 1] != e[i].id);
            }

        } else if (DFN[cur] > DFN[v]) {
            low[cur] = min(low[cur], DFN[v]);
            st[++top] = e[i].id;
        }
    }
}
void solveTarjan(int n) {
    memset(DFN, 0, sizeof DFN);
    memset(low, 0, sizeof low);
    memset(iscut, 0, sizeof iscut);
    memset(id, 0, sizeof id);
    for (int i = 1; i <= maxm - 2; ++i) {
        bolg[i].clear();
    }
    when = top = toSelid = 0;
    for (int i = 1; i <= n; ++i) {
        if (!DFN[i]) {
            root = i;
            tarjan(i, i);
        }
    }
}
int dis[maxm];
bool treeCut[maxm], vis[maxm];
struct Node {
    int cur, cnt;
    Node(int _cur, int _cnt) {
        cur = _cur, cnt = _cnt;
    }
};
void bfs(int be) {
    vis[be] = true;
    dis[be] = treeCut[be];
    queue<struct Node>que;
    que.push(Node(be, treeCut[be]));
    while (!que.empty()) {
        struct Node t = que.front();
        que.pop();
        for (int i = first[t.cur]; i; i = e[i].tonext) {
            int v = e[i].v;
            if (vis[v]) continue;
            vis[v] = true;
            dis[v] = t.cnt + treeCut[v];
            que.push(Node(v, t.cnt + treeCut[v]));
        }
    }
}
int ansc[maxn * 4][22], deep[maxm], fa[maxm];
void init_LCA(int cur) {
    ansc[cur][0] = fa[cur]; //跳1步，那么祖先就是爸爸
    for (int i = 1; i <= 20; ++i) { //倍增思路，递归处理
        ansc[cur][i] = ansc[ansc[cur][i - 1]][i - 1];
    }
    for (int i = first[cur]; i; i = e[i].tonext) {
        int v = e[i].v;
        if (v == fa[cur]) continue;
        fa[v] = cur;
        deep[v] = deep[cur] + 1;
        init_LCA(v);
    }
}
int LCA(int x, int y) {
    if (deep[x] < deep[y]) swap(x, y); //需要x是最深的
    for (int i = 20; i >= 0; --i) { //从大到小枚举，因为小的更灵活
        if (deep[ansc[x][i]] >= deep[y]) { //深度相同，走进去就对了。就是要去到相等。
            x = ansc[x][i];
        }
    }
    if (x == y) return x;
    for (int i = 20; i >= 0; --i) {
        if (ansc[x][i] != ansc[y][i]) { //走到第一个不等的地方，
            x = ansc[x][i];
            y = ansc[y][i];
        }
    }
    return ansc[x][0]; //再跳一步就是答案
}
void init() {
    num = 0;
    memset(ansc, 0, sizeof ansc);
    memset(deep, 0, sizeof deep);
    memset(fa, 0, sizeof fa);
    memset(dis, 0, sizeof dis);
    memset(treeCut, 0, sizeof treeCut);
    memset(vis, 0, sizeof vis);

}
void work() {
    init();
    num = 0;
    memset(first, 0, sizeof first);
    for (int i = 1; i <= m; ++i) {
        int u, v;
        scanf("%d%d", &u, &v);
        addEdge(u, v, i);
        addEdge(v, u, i);
        uuu[i] = u;
        vvv[i] = v;
    }
    solveTarjan(n);
    memset(first, 0, sizeof first);
    memset(treeCut, 0, sizeof treeCut);
    num = 0;
    int to = toSelid;
    for (int i = 1; i <= n; ++i) {
        if (!iscut[i]) continue;
        ++to;
        treeCut[to] = true;
        sort(bolg[i].begin(), bolg[i].end());
        addEdge(to, bolg[i][0], 1);
        addEdge(bolg[i][0], to, 1);
        for (int j = 1; j < bolg[i].size(); ++j) {
            if (bolg[i][j - 1] == bolg[i][j]) continue;
            addEdge(to, bolg[i][j], 1);
            addEdge(bolg[i][j], to, 1);
        }
    }
//    int tot = 2;
//    for (int i = 0;  i < bolg[tot].size(); ++i) {
//        cout << bolg[tot][i] << " ";
//    }
//    cout << endl;
    memset(vis, false, sizeof vis);
    memset(fa, 0, sizeof fa);
    memset(deep, 0, sizeof deep);
    memset(ansc, 0, sizeof ansc);
    memset(dis, 0, sizeof dis);
    for (int i = 1; i <= to; ++i) {
        if (vis[i]) continue;
        bfs(i);
        fa[i] = i;
        deep[i] = 0;
        init_LCA(i);
    }
//    cout << iscut[11] << " ***" << endl;
    int q;
    scanf("%d", &q);
    while (q--) {
        int x, y;
        scanf("%d%d", &x, &y);
        x = id[x];
        y = id[y];
        if (x == y)  {
            printf("0
");
            continue;
        }
        int res = LCA(x, y);
        if (res == x) {
            assert(treeCut[x] == 0);
        }
        if (res == y) {
            assert(treeCut[y] == 0);
        }
        int ans = dis[x] + dis[y] - 2 * dis[res];
        assert(ans + treeCut[res] >= 0);
        printf("%d
", ans + treeCut[res]);
    }
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    while (scanf("%d%d", &n, &m) != EOF && (n + m)) work();
    return 0;
}

原来这题是有重边的，怪不得我一直wa。而且自己生成的数据都是没有重边的，TAT

说下思路，

题目要求的经过多少个割点，

首先，题目给出的是从边s走到边t。这样很好办，一开始以为是点s做到点t，这样比较麻烦。

先做一次点双连通分量，对边分好id。然后对于每一个割点，把和它连接的边重建一颗树。然后这个割点就是这棵树的treeCut

这就相当于给出一颗树，树上有些点（那副图的边变成点了）是染色了的，问从点s走到点t，最少走过多少个染色的点。

这样可以预处理出dis[i]表示从树根走到点i，一共经过多少个染色的点，这样就相当于区间减法，然后配合lca即可。

1、题目虽然保证roads到rodat有路，但是可能有多个连通分支。

2、有重边。

其实有重边不用怕，没什么的，只需要标记那条边是用过了的就可以。

既可以判断不返回fa，又可以判重边。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;


#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
const int maxm = 200000 + 20;
const int maxn = 10000 + 20;
struct Edge {
    int u, v, id, tonext;
} e[maxm * 2];
int first[maxm], num;
void addEdge(int u, int v, int id) {
    ++num;
    e[num].u = u, e[num].v = v, e[num].tonext = first[u];
    first[u] = num;
    e[num].id = id;
}
int uuu[maxm], vvv[maxm];
int n, m;
int low[maxm], DFN[maxm], st[maxm], id[maxm];
int top, when, toSelid;
bool iscut[maxm];
vector<int>bolg[maxm];
int root;
void tarjan(int cur, int fa, int fromID) {
    DFN[cur] = low[cur] = ++when;
    int child = 0;
    for (int i = first[cur]; i; i = e[i].tonext) {
        int v = e[i].v;
        if (v == fa && e[i].id == fromID) continue;
        if (!DFN[v]) {
            child++;
            st[++top] = e[i].id;
            tarjan(v, cur, e[i].id);
            low[cur] = min(low[cur], low[v]);
            if (low[v] >= DFN[cur]) {
                iscut[cur] = true;
//                if (cur == root && child < 2) iscut[cur] = false;
                ++toSelid;
                do {
                    int eID = st[top--];
                    bolg[uuu[eID]].push_back(toSelid);
                    bolg[vvv[eID]].push_back(toSelid);
                    id[eID] = toSelid;
                } while (st[top + 1] != e[i].id);
            }

        } else if (DFN[cur] > DFN[v]) {
            low[cur] = min(low[cur], DFN[v]);
            st[++top] = e[i].id;
        }
    }
}
void solveTarjan(int n) {
    memset(DFN, 0, sizeof DFN);
    memset(low, 0, sizeof low);
    memset(iscut, 0, sizeof iscut);
    memset(id, 0, sizeof id);
    for (int i = 1; i <= maxm - 2; ++i) {
        bolg[i].clear();
    }
    when = top = toSelid = 0;
    for (int i = 1; i <= n; ++i) {
        if (!DFN[i]) {
            root = i;
            tarjan(i, i, -1);
        }
    }
}
int dis[maxm];
bool treeCut[maxm], vis[maxm];
struct Node {
    int cur, cnt;
    Node(int _cur, int _cnt) {
        cur = _cur, cnt = _cnt;
    }
};
void bfs(int be) {
    vis[be] = true;
    dis[be] = treeCut[be];
    queue<struct Node>que;
    que.push(Node(be, treeCut[be]));
    while (!que.empty()) {
        struct Node t = que.front();
        que.pop();
        for (int i = first[t.cur]; i; i = e[i].tonext) {
            int v = e[i].v;
            if (vis[v]) continue;
            vis[v] = true;
            dis[v] = t.cnt + treeCut[v];
            que.push(Node(v, t.cnt + treeCut[v]));
        }
    }
}
int ansc[maxn * 4][22], deep[maxm], fa[maxm];
void init_LCA(int cur) {
    ansc[cur][0] = fa[cur]; //跳1步，那么祖先就是爸爸
    for (int i = 1; i <= 20; ++i) { //倍增思路，递归处理
        ansc[cur][i] = ansc[ansc[cur][i - 1]][i - 1];
    }
    for (int i = first[cur]; i; i = e[i].tonext) {
        int v = e[i].v;
        if (v == fa[cur]) continue;
        fa[v] = cur;
        deep[v] = deep[cur] + 1;
        init_LCA(v);
    }
}
int LCA(int x, int y) {
    if (deep[x] < deep[y]) swap(x, y); //需要x是最深的
    for (int i = 20; i >= 0; --i) { //从大到小枚举，因为小的更灵活
        if (deep[ansc[x][i]] >= deep[y]) { //深度相同，走进去就对了。就是要去到相等。
            x = ansc[x][i];
        }
    }
    if (x == y) return x;
    for (int i = 20; i >= 0; --i) {
        if (ansc[x][i] != ansc[y][i]) { //走到第一个不等的地方，
            x = ansc[x][i];
            y = ansc[y][i];
        }
    }
    return ansc[x][0]; //再跳一步就是答案
}
void work() {
    num = 0;
    memset(first, 0, sizeof first);
    for (int i = 1; i <= m; ++i) {
        int u, v;
        scanf("%d%d", &u, &v);
        addEdge(u, v, i);
        addEdge(v, u, i);
        uuu[i] = u;
        vvv[i] = v;
    }
    solveTarjan(n);
    memset(first, 0, sizeof first);
    memset(treeCut, 0, sizeof treeCut);
    num = 0;
    int to = toSelid;
    for (int i = 1; i <= n; ++i) {
        if (!iscut[i]) continue;
        ++to;
        treeCut[to] = true;
        sort(bolg[i].begin(), bolg[i].end());
        addEdge(to, bolg[i][0], 1);
        addEdge(bolg[i][0], to, 1);
        for (int j = 1; j < bolg[i].size(); ++j) {
            if (bolg[i][j - 1] == bolg[i][j]) continue;
            addEdge(to, bolg[i][j], 1);
            addEdge(bolg[i][j], to, 1);
        }
    }
//    int tot = 2;
//    for (int i = 0;  i < bolg[tot].size(); ++i) {
//        cout << bolg[tot][i] << " ";
//    }
//    cout << endl;
    memset(vis, false, sizeof vis);
    for (int i = 1; i <= to; ++i) {
        if (vis[i]) continue;
        bfs(i);
        fa[i] = i;
        deep[i] = 0;
        init_LCA(i);
    }
//    cout << iscut[11] << " ***" << endl;
    int q;
    scanf("%d", &q);
    while (q--) {
        int x, y;
        scanf("%d%d", &x, &y);
        x = id[x];
        y = id[y];
        if (x == y)  {
            printf("0
");
            continue;
        }
        int res = LCA(x, y);
        if (res == x) {
            assert(treeCut[x] == 0);
        }
        if (res == y) {
            assert(treeCut[y] == 0);
        }
        int ans = dis[x] + dis[y] - 2 * dis[res];
        assert(ans + treeCut[res] >= 0);
        printf("%d
", ans + treeCut[res]);
    }
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    while (scanf("%d%d", &n, &m) != EOF && (n + m)) work();
    return 0;
}

代码写的很烂，因为wa了3小时，一直在yy改错。