tarjan算法c语言,tarjan算法板子 - osc_e45irv7l的个人空间 - OSCHINA - 中文开源技术交流社区...
无向图
概念
时间戳
\(dfn[x]\),在深度优先遍历中，按照每个节点第一次被访问的顺序，依次做整数标记
追溯值
\(low[x]\),通过非搜索边能到达的最小时间戳
割边判定法则
无向边\((x,y)\)是割边/桥，当且仅当存在x的一个子节点满足\(dfn[x] < low[y]\)
删除无向边\((x,y)\)后，图断开成两个部分
板子
int dfn[N], low[N], dfcnt;
bool g[M];
void tarjan(int x, int ei) {
dfn[x] = low[x] = ++dfcnt;
for(int i = head[x]; i; i = e[i].next) {
int y = e[i].t;
if (!dfn[y]) {
tarjan(y, i);
low[x] = min(low[x], low[y]);
if (dfn[x] < low[y]) g[i] = g[i^1] = 1;
}
else if (i != (ei^1)) low[x] = min(low[x], dfn[y]);
}
}
割点判定法则
若x不是根节点，则x是割点当且仅当存在一个子节点y满足\(dfn[x]\leq low[y]\)
若x是根节点，则x是割点当且仅当存在至少两个子节点\(y_1,y_2\)满足上条件
板子
int dfn[N], low[N], dfcnt, rt;
bool g[N];
void tarjan(int x) {
dfn[x] = low[x] = ++dfcnt;
int son = 0;
for (int i = head[x]; i; i = e[i].next) {
int y = e[i].t;
if (!dfn[y]) {
tarjan(y);
low[x] = min(low[x], low[y]);
if (dfn[x] <= low[y]) {
son++;
if (x != rt || son > 1) g[x] = 1;
}
}
else low[x] = min(low[x], dfn[y]);
}
}
点双联通分量
对于，每个点双中来说，图里是不存在割点的
板子
int dfn[N], low[N], dfcnt, sta[N], top, cnt;
vector dcc[N];
bool g[N];
void tarjan(int x, int rt) {
dfn[x] = low[x] = ++dfcnt;
sta[++top] = x;
int son = 0;
for (int i = head[x]; i; i = e[i].next) {
int y = e[i].t;
if (!dfn[y]) {
tarjan(y);
low[x] = min(low[x], low[y]);
if (dfn[x] <= low[y]) {
son++;
if (x != rt || son > 1) g[x] = 1;
dcc[++cnt].clear();
while (1) {
int z = sta[top--];
dcc[cnt].push_back(z);
if (y == z) break;
}
dcc[cnt].push_back(x);
}
}
else low[x] = min(low[x], dfn[y]);
}
}
边双联通分量
对于一个边双，任意两个点都有两条不重合的路径
板子
int dfn[N], low[N], dfcnt;
bool g[M];
void tarjan(int x, int ei) {
dfn[x] = low[x] = ++dfcnt;
for(int i = head[x]; i; i = e[i].next) {
int y = e[i].t;
if (!dfn[y]) {
tarjan(y, i);
low[x] = min(low[x], low[y]);
if (dfn[x] < low[y]) g[i] = g[i^1] = 1;
}
else if (i != (ei^1)) low[x] = min(low[x], dfn[y]);
}
}
int n, m, d[N], b[N], cnt, ans;
void dfs(int x) {
b[x] = cnt;
for(int i = head[x]; i; i = e[i].next) {
int y = e[i].t;
if (b[y] || g[i]) continue;
dfs(y);
}
}
int main() {
//~~~
for(int i = 1; i <= n; i++)
if (!dfn[i]) tarjan(i, 0);
for(int i = 1; i <= n; i++)
if (!b[i]) cnt++, dfs(i);
//~~~
return 0;
}
有向图
有向图的强联通分量
在一个强联通分量中，存在x到y的路径，就存在y到x的路径
板子
void tarjan(int x) {
dfn[x] = low[x] = ++dfcnt;
s[++top] = x;
for(int i = head[x]; i; i = e[i].next) {
int y = e[i].t;
if (!dfn[y]) tarjan(y), low[x] = min(low[x], low[y]);
else if (!b[y]) low[x] = min(low[x], dfn[y]);
}
if (dfn[x] == low[x]) {
cnt++;
while(1) {
int y = s[top--];
b[y] = cnt;
size[cnt]++;
if (x == y) break;
}
}
}
例题
luoguP3387缩点
代码
#include
#include
#include
#include
using namespace std;
const int N = 1e4+5, M = 1e5+5;
struct side { int t, next; } e[M][2];
int head[N][2], tot[2];
void add(int x, int y, int k) {
e[++tot[k]][k].next = head[x][k];
head[x][k] = tot[k];
e[tot[k]][k].t = y;
}
int n, m, w[N], r[N], d[N], ans;
int dfn[N], low[N], dfcnt, sta[N], top, cnt, bel[N], sum[N];
void tarjan(int x) {
dfn[x] = low[x] = ++dfcnt;
sta[++top] = x;
for (int i = head[x][0]; i; i = e[i][0].next) {
int y = e[i][0].t;
if (!dfn[y]) tarjan(y), low[x] = min(low[x], low[y]);
else if (!bel[y]) low[x] = min(low[x], dfn[y]);
}
if (dfn[x] == low[x]) {
cnt++;
while (1) {
int y = sta[top--];
bel[y] = cnt;
sum[cnt] += w[y];
if (x == y) break;
}
}
}
queue q;
int tuopu() {
for (int i = 1; i <= cnt; i++)
if (!r[i]) q.push(i), d[i] = sum[i];
while (!q.empty()) {
int x = q.front(); q.pop();
for (int i = head[x][1]; i; i = e[i][1].next) {
int y = e[i][1].t;
d[y] = max(d[y], d[x] + sum[y]);
if (--r[y] == 0) q.push(y);
}
}
for (int i = 1; i <= cnt; i++)
ans = max(ans, d[i]);
return ans;
}
int main() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++)
scanf("%d", &w[i]);
for (int i = 1; i <= m; i++) {
int x, y;
scanf("%d%d", &x, &y);
add(x, y, 0);
}
for (int i = 1; i <= n; i++)
if (!dfn[i]) tarjan(i);
for (int x = 1; x <= n; x++)
for (int i = head[x][0]; i; i = e[i][0].next) {
int y = e[i][0].t;
if (bel[x] != bel[y])
r[bel[y]]++, add(bel[x], bel[y], 1);
}
printf("%d\n", tuopu());
return 0;
}
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