一、内容
N planets are connected by M bidirectional channels that allow instant transportation. It's always possible to travel between any two planets through these channels.
If we can isolate some planets from others by breaking only one channel , the channel is called a bridge of the transportation system.
People don't like to be isolated. So they ask what's the minimal number of bridges they can have if they decide to build a new channel.
Note that there could be more than one channel between two planets.
Input
The input contains multiple cases.
Each case starts with two positive integers N and M , indicating the number of planets and the number of channels.
(2<=N<=200000, 1<=M<=1000000)
Next M lines each contains two positive integers A and B, indicating a channel between planet A and B in the system. Planets are numbered by 1..N.
A line with two integers '0' terminates the input.
Output
For each case, output the minimal number of bridges after building a new channel in a line.
Sample Input
4 4
1 2
1 3
1 4
2 3
0 0
Sample Output
0
二、思路
- DCC缩点后建立一棵树， 树中的所有边都是桥。 任意连接一条边使桥的数量最少。 那么就是要找出树的最长路径–树的直径。
- ans = 桥的数量 - 树的直径
三、代码
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int N = 2e5 + 5, M = 4e6 + 6;
struct E {int v, next;} e[M];
int n, m, u, v, len, ans, dh[N], h[N], dcc_cnt, id[N], num, dfn[N], low[N], d[N];
bool brid[M], vis[N];
void add(int h[], int u, int v) {e[++len].v = v; e[len].next = h[u]; h[u] = len;}
void tarjan(int u, int in_edge) {
dfn[u] = low[u] = ++num;
for (int j = h[u]; j; j = e[j].next) {
int v = e[j].v;
if (!dfn[v]) {
tarjan(v, j);
low[u] = min(low[u], low[v]);
if (dfn[u] < low[v]) brid[j] = brid[j ^ 1] = true;
} else if ((j ^ 1) != in_edge) low[u] = min(low[u], dfn[v]);
}
}
void dfs(int u) {
id[u] = dcc_cnt;
for (int j = h[u]; j; j = e[j].next) {
int v = e[j].v;
if (id[v] || brid[j]) continue;
dfs(v);
}
}
void dp(int u) {
vis[u] = true;
for (int j = dh[u]; j; j = e[j].next) {
int v = e[j].v;
if (vis[v]) continue;
dp(v);
ans = max(ans, d[u] + d[v] + 1);
d[u] = max(d[u], d[v] + 1);
}
}
int main() {
while (scanf("%d%d", &n, &m), n) {
memset(h, 0, sizeof(h)); len = num = 1;
memset(dh, 0, sizeof(dh));
memset(id, 0, sizeof(id));
memset(dfn, 0, sizeof(dfn));
memset(d, 0, sizeof(d));
memset(brid, false, sizeof(brid));
memset(vis, false, sizeof(vis));
for (int i = 1; i <= m; i++) {
scanf("%d%d", &u, &v);
add(h, u, v); add(h, v, u);
}
tarjan(1, 0); dcc_cnt = 0;
//缩点 建树
for (int i = 1; i <= n; i++) {
if (!id[i]) {
dcc_cnt++;
dfs(i);
}
}
int tlen = len;
for (int j = 2; j <= tlen; j += 2) {
u = id[e[j].v], v = id[e[j ^ 1].v];
if (u == v) continue;
add(dh, u, v); add(dh, v, u);
}
ans = 0;
dp(1); //求出直径
printf("%d\n", dcc_cnt - 1 - ans);
}
return 0;
}