POJ3694-Network(Tarjan缩点+LCA)

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题目链接

题意：给你一个连通图，然后再给你n个询问，每个询问给一个点u,v表示加上u,v之后又多少个桥。

思路：用Tarjan缩点后，形成一棵树，所以树边都是桥了。然后加入边以后，查询LCA，LCA上的桥都减掉。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <utility>#include <algorithm>using namespace std;const int MAXN = 100005;struct Edge{    int to, next;}edge[MAXN * 5];int head[MAXN], tot;int Low[MAXN], DFN[MAXN];int Index, top;int bridge;int n, m;int f[MAXN], ancestor[MAXN];int find(int x) {    return x == f[x] ? x : f[x] = find(f[x]);}bool Union(int a, int b) {    int pa = find(a);    int pb = find(b);    if (pa != pb) {        f[pa] = pb;        return true;    }    return false;}void addedge(int u, int v) {    edge[tot].to = v;    edge[tot].next = head[u];    head[u] = tot++;}void Tarjan(int u, int pre) {    int v;    Low[u] = DFN[u] = ++Index;    for (int i = head[u]; i != -1; i = edge[i].next) {        v = edge[i].to;        if (v == pre) continue;         if (!DFN[v]) {            Tarjan(v, u);             ancestor[v] = u;            if (Low[u] > Low[v]) Low[u] = Low[v];            if (Low[v] > DFN[u])                 bridge++;              else                 Union(u, v);        }         else if (Low[u] > DFN[v])            Low[u] = DFN[v];    }}void init() {    memset(head, -1, sizeof(head));    memset(DFN, 0, sizeof(DFN));     tot = 0;    Index = top = 0;    bridge = 0;    memset(ancestor, 0, sizeof(ancestor));    for (int i = 1; i <= n; i++)        f[i] = i;}void solve() {    for (int i = 1; i <= n; i++)        if (!DFN[i])            Tarjan(i, -1);}void lca(int u, int v) {    while (u != v) {        while (DFN[u] >= DFN[v] && u != v) {            if (Union(u, ancestor[u]))                  bridge--;            u = ancestor[u];        }        while (DFN[v] >= DFN[u] && u != v) {            if (Union(v, ancestor[v]))                  bridge--;            v = ancestor[v];        }    }}int main() {    int t = 1;    while (scanf("%d%d", &n, &m)) {        if (n == 0 && m == 0) break;        init();         int u, v;          for (int i = 0; i < m; i++) {            scanf("%d%d", &u, &v);             addedge(u, v);            addedge(v, u);        }        solve();        printf("Case %d:\n", t++);        int k;        scanf("%d", &k);        while (k--) {            scanf("%d%d", &u, &v);             lca(u, v);            printf("%d\n", bridge);        }        printf("\n");    }    return 0;}

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