LOJ 10162 骑士
把不能在一起的骑士连一条边，可以发现每个连通块最多一个环。
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59#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e6 + 4;
int n, a[N], b[N], cnt = 1, head[N], x1, x2, now, vis[N];
ll f[N][2];
struct E {int nxt, v;} e[N << 1];
void add(int x, int y) {
e[++cnt] = {head[x], y}; head[x] = cnt;
e[++cnt] = {head[y], x}; head[y] = cnt;
}
void find(int u, int fr) {
vis[u] = 1;
for (int i = head[u]; i; i = e[i].nxt) {
int v = e[i].v;
if ((i ^ 1) == fr) continue;
if (vis[v]) {
now = i, x1 = u, x2 = v;
continue;
}
find(v, i);
}
}
void dfs(int u, int pre) {
f[u][1] = a[u];
f[u][0] = 0;
for (int i = head[u]; i; i = e[i].nxt) {
if (i == now || (i ^ 1) == now || (i ^ 1) == pre) continue;
dfs(e[i].v, i);
f[u][0] = f[u][0] + max(f[e[i].v][0], f[e[i].v][1]);
f[u][1] = f[u][1] + f[e[i].v][0];
}
}
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
int v;
scanf("%d %d", &a[i], &v);
add(i, v);
}
ll res = 0;
for (int i = 1; i <= n; i++) {
ll ans = 0;
if (!vis[i]) {
find(i, -2);
dfs(x1, -2);
ans = f[x1][0];
dfs(x2, -2);
res += max(ans, f[x2][0]);
}
}
printf("%lld\n", res);
return 0;
}