给定无向连通图G=(V,E);
割点:
若对于,从图中删去节点x与所有与x关联的边后,G分为两个或以上不相连的子图,则称x为G的割点。
割边:
若对于,从图中删去边e之后,G分为两个不相连的子图,则称e为桥或割边。
割边的判定法则:
无向边(x,y)是桥,当且仅当搜索树上存在x的一个子节点y,满足:dfn[x]
hdu4738
#include
using namespace std;
const int N = 2e6 + 10;
int n, m, cnt, h[N], dfn[1010], low[1010], ind, ans, mp[1010][1010];
struct node {
int v, nt, w;
} no[N];
void add(int u, int v, int w) {
no[cnt] = node{v, h[u], w};
h[u] = cnt++;
}
void tarjan(int u, int fa) {
low[u] = dfn[u] = ++ind;
for(int i = h[u]; ~i; i = no[i].nt) {
int v = no[i].v;
if(!dfn[v]) {
tarjan(v, u);
low[u] = min(low[u], low[v]);
if(dfn[u] < low[v])
ans = min(ans, mp[u][v]);//最小边
} else if(v != fa)
low[u] = min(low[u], dfn[v]);
}
}
int main() {
while(~scanf("%d%d", &n, &m) && (n + m)) {
memset(h, -1, sizeof h), cnt = 0, ind = 0, ans = 10010;
memset(dfn, 0, sizeof dfn);
memset(low, 0, sizeof low);
memset(mp, 0, sizeof mp);
for(int u, v, w, i = 1; i <= m; i++) {
scanf("%d%d%d", &u, &v, &w);
if(mp[u][v] == 0)//处理重边
add(u, v, w), add(v, u, w);
else if(mp[u][v])
mp[u][v] = mp[v][u] = 10010;
if(mp[v][u] == 0)
mp[u][v] = mp[v][u] = w;
}
int lt = 0;
for(int i = 1; i <= n; i++)
if(!dfn[i])
tarjan(i, 0), lt++;
if(ans == 10010)
ans = -1;
if(ans == 0)
ans = 1;
if(lt > 1)
ans = 0;
printf("%d\n", ans);
}
return 0;
}
割点判定法则:
若x不是搜索树的根节点(深度优先遍历的起点),则x是割点当且仅当搜索树上存在x的一个子节点y,满足:dfn[x]low[y]
求割点模板:
#include
#define ll long long
using namespace std;
const int N = 1e6 + 10;
int n, m, h[N], cnt, dfn[N], low[N], ans, isc[N], ind;
struct node {
int v, nt;
} no[N];
void add(int u, int v) {
no[cnt].v = v;
no[cnt].nt = h[u];
h[u] = cnt++;
}
void tarjan(int u, int root) {
dfn[u] = low[u] = ++ind;
int sun = 0;
for(int i = h[u]; ~i; i = no[i].nt) {
int v = no[i].v;
if(!dfn[v]) {
tarjan(v, root);//回溯时求
low[u] = min(low[u], low[v]);
if(u == root)
sun++;
if(low[v] >= dfn[u] && u != root)//形成环
isc[u] = 1;
}
low[u] = min(low[u], dfn[v]);
}
if(sun >= 2 && u == root)//有两颗以上的子树
isc[root] = 1;
}
int main() {
memset(h, -1, sizeof h);
scanf("%d%d", &n, &m);
for(int u, v, i = 1; i <= m; i++) {
scanf("%d%d", &u, &v);
add(u, v), add(v, u);
}
for(int i = 1; i <= n; i++)
if(!dfn[i])
tarjan(i, i);
for(int i = 1; i <= n; i++)
if(isc[i])
ans++;
printf("%d\n", ans);
for(int i = 1; i <= n; i++)
if(isc[i])
printf("%d ", i);
return 0;
}