SCC
缩点
原理
见《进阶指南》第415
页。
强连通分量编号递减的序列一定满足拓扑序。
模板题
#include <iostream>
#include <cstring>
#include <queue>
#include <stack>
using namespace std;
const int N = 10010, M = 100010;
struct edge {
int to, next;
};
edge e[M];
int idx, head[N], in_deg[N];
struct node {
int u, v;
};
int n, w1[N], m;
node rec[M];
int cnt, dfn[N], low[N];
stack<int> s;
bool in_stack[N];
int tot, scc[N], w2[N];
int f[N];
void add_edge (int u, int v) {
e[idx].to = v;
e[idx].next = head[u];
head[u] = idx ++;
}
void tarjan (int cur) {
low[cur] = dfn[cur] = ++ cnt;
s.push(cur);
in_stack[cur] = true;
for (int i = head[cur]; i != -1; i = e[i].next) {
int to = e[i].to;
if (dfn[to] == 0) {
tarjan(to);
low[cur] = min(low[cur], low[to]);
} else if (in_stack[to] == true) {
low[cur] = min(low[cur], dfn[to]);
}
}
if (dfn[cur] == low[cur]) {
++ tot;
int t;
do {
t = s.top();
s.pop();
in_stack[t] = false;
scc[t] = tot;
w2[tot] += w1[t];
} while (t != cur);
}
}
void topo_sort () {
queue<int> q;
for (int i = 1; i <= tot; ++ i) {
if (in_deg[i] == 0) {
q.push(i);
f[i] = w2[i];
}
}
while (!q.empty()) {
int cur = q.front();
q.pop();
for (int i = head[cur]; i != -1; i = e[i].next) {
int to = e[i].to;
f[to] = max(f[to], f[cur] + w2[to]);
if (-- in_deg[to] == 0) q.push(to);
}
}
}
int main () {
memset(head, -1, sizeof(head));
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; ++ i) scanf("%d", &w1[i]);
for (int i = 1; i <= m; ++ i) {
scanf("%d%d", &rec[i].u, &rec[i].v);
add_edge(rec[i].u, rec[i].v);
}
for (int i = 1; i <= n; ++ i)
if (dfn[i] == 0)
tarjan(i);
idx = 0;
memset(head, -1, sizeof(head));
for (int i = 1; i <= m; ++ i) {
int u = rec[i].u, v = rec[i].v;
if (scc[u] == scc[v]) continue;
add_edge(scc[u], scc[v]);
++ in_deg[scc[v]];
}
topo_sort();
int res = 0;
for (int i = 1; i <= tot; ++ i)
res = max(res, f[i]);
printf("%d", res);
return 0;
}