bzoj4316: 小C的独立集
链接
思路
不是环的边==没有上司的舞会。
其他的，把环拿出来,考虑与深度最小的点u的交界处的点选不选，进行两次dp更新f[u]
代码
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 7;
int read() {
int x = 0, f = 1; char s = getchar();
for (; s > '9' || s < '0'; s = getchar()) if (s == '-') f = -1;
for (; s >= '0' && s <= '9'; s = getchar()) x = x * 10 + s - '0';
return x * f;
}
int n, m;
struct node {
int v, nxt;
}e[N << 1];
int head[N << 1], tot;
void add(int u, int v) {
e[++tot].v = v ;
e[tot].nxt = head[u];
head[u] = tot;
}
int low[N], dfn[N], cnt, fa[N], f[N][2] ;
void work(int u, int y) {
int t0, t1, f0=0, f1=0;
for(int i = y;i != u;i = fa[i]){
t0 = f0 + f[i][0], t1 = f1 + f[i][1];
f0 = max(t0, t1), f1 = t0;
}
f[u][0] += f0;
f0 = 0, f1 = -1e9;
for(int i = y;i != u;i = fa[i]){
t0 = f0 + f[i][0], t1 = f1 + f[i][1];
f0 = max(t0, t1), f1 = t0;
}
f[u][1] += f1;
}
void tarjan(int u, int father) {
fa[u] = father;
low[u] = dfn[u] = ++cnt;
f[u][1] = 1, f[u][0] = 0;
for (int i = head[u]; i; i = e[i].nxt) {
int v = e[i].v;
if (!dfn[v]) {
tarjan(v, u);
low[u] = min(low[u], low[v]);
} else if (v != father) low[u] = min(low[u], dfn[v]);
if (dfn[u] < low[v]) {
f[u][0] += max(f[v][1], f[v][0]);
f[u][1] += f[v][0];
}
}
for (int i = head[u]; i; i = e[i].nxt) {
int v = e[i].v;
if(u != fa[v] && dfn[v] > dfn[u])
work(u, v);
}
}
int main() {
n = read(), m = read();
for (int i = 0; i < m; ++i) {
int u = read(), v = read();
add(u, v), add(v, u);
}
tarjan(1, 0);
printf("%d
", max(f[1][0], f[1][1]));
return 0;
}