POJ - 3352 Road Construction(边双连通分量)
题目大意:给出一张无向图。问加入多少边才干使得这张无向图变成边双连通分量
解题思路:先求出全部的边双连通分量。再将边双连通缩成一个点。通过桥连接起来。这样就形成了一棵无根树了
如今的问题是,将这颗无根树变成边双连通分量
网上的解释是:统计出树中度为1的节点的个数,即为叶节点的个数,记为leaf。则至少在树上加入(leaf+1)/2条边,就能使树达到边二连通,所以至少加入的边数就是(leaf+1)/2。具体方法为。首先把两个近期公共祖先最远的两个叶节点之间连接一条边,这样能够把这两个点到祖先的路径上全部点收缩到一起,由于一个形成的环一定是双连通的。
然后再找两个近期公共祖先最远的两个叶节点,这样一对一对找完,恰好是(leaf+1)/2次,把全部点收缩到了一起。
#include <cstdio>
#include <cstring>
#define N 1010
#define min(a,b) ((a)<(b) ?(a):(b))
struct Edge{
int to, next;
}E[N*2];
int n, m, tot, dfs_clock, bcc_cnt, top, bnum;;
int head[N], pre[N], belong[N], degree[N], stack[N], bridge[N][2];
void Addegreedge(int u, int v) {
E[tot].to = v; E[tot].next = head[u]; head[u] = tot++;
u = u ^ v; v = v ^ u; u = u ^ v;
E[tot].to = v; E[tot].next = head[u]; head[u] = tot++;
}
void init() {
memset(head, -1, sizeof(head));
tot = 0;
int u, v;
for (int i = 0; i < m; i++) {
scanf("%d%d", &u, &v);
Addegreedge(u, v);
}
}
int dfs(int u, int fa) {
int lowu = pre[u] = ++dfs_clock;
stack[++top] = u;
for (int i = head[u]; i != -1; i = E[i].next) {
int v = E[i].to;
if (!pre[v]) {
int lowv = dfs(v, u);
lowu = min(lowv, lowu);
if (lowv > pre[u]) {
bridge[bnum][0] = u;
bridge[bnum++][1] = v;
bcc_cnt++;
while (1) {
int x = stack[top--];
belong[x] = bcc_cnt;
if (x == v) break;
}
}
}else if (pre[v] < pre[u] && v != fa) {
lowu = min(lowu, pre[v]);
}
}
return lowu;
}
void solve() {
memset(degree, 0, sizeof(degree));
memset(pre, 0, sizeof(pre));
dfs_clock = bcc_cnt = top = bnum = 0;
dfs(1, -1);
if (top) {
bcc_cnt++;
while (1) {
int x = stack[top--];
belong[x] = bcc_cnt;
if (x == 1)
break;
}
}
for (int i = 0; i < bnum; i++) {
int u = bridge[i][0];
int v = bridge[i][1];
degree[belong[u]]++;
degree[belong[v]]++;
}
int leaf = 0;
for (int i = 1; i <= bcc_cnt; i++)
if (degree[i] == 1)
leaf++;
printf("%d\n", (leaf + 1)/ 2);
}
int main() {
while (scanf("%d%d", &n, &m) != EOF) {
init();
solve();
}
return 0;
}