bzoj4316: 小C的獨(dú)立集

鏈接

bzoj

思路

不是環(huán)的邊==沒有上司的舞會(huì)。
其他的，把環(huán)拿出來,考慮與深度最小的點(diǎn)u的交界處的點(diǎn)選不選，進(jìn)行兩次dp更新f[u]

代碼

#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\n", max(f[1][0], f[1][1]));return 0; }

轉(zhuǎn)載于:https://www.cnblogs.com/dsrdsr/p/10957520.html

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