无归岛[HNOI2009]

题目描述

https://www.luogu.com.cn/problem/P4410

题解

原图显然是一个仙人掌(似乎还有些别的性质 但是其实没什么必要)

先考虑树的情况，题意即为不能同时选择相邻的两点，设 \(f[x][0/1]\) 表示选/不选 \(x\) 时， \(x\) 子树内的最大战斗力

\(f[x][0]=\max\limits_{y\in son(x)} (\max(f[y][0],f[y][1]))\)

\(f[x][1]=\max\limits_{y\in son(x)} (f[y][0]) + A_x\)

在环上如何dp？

在环上找到一个点 \(x\) 并从那里把环断开，分 \(x\) 选或 \(x\) 不选两种，分别进行dp来推出 \(f[x][0/1]\) 即可

代码

#include <bits/stdc++.h>
#define N 500005
using namespace std;

int n, m, ans, a[N], f[N][2];
int head[N], pre[N<<1], to[N<<1], sz;

inline void addedge(int u, int v) {
	pre[++sz] = head[u]; head[u] = sz; to[sz] = v;
	pre[++sz] = head[v]; head[v] = sz; to[sz] = u;
}

int dfn[N], low[N], stk[N], top, c[N], tot, tme;

void solve(int x) {
	int a0 = 0, a1 = 0, b0 = 0, b1 = -0x3f3f3f3f; //不选x
	for (int i = 1; i <= tot; i++) {
		a0 = max(b0, b1) + f[c[i]][0]; a1 = b0 + f[c[i]][1];
		b0 = a0; b1 = a1;
	}
	f[x][0] += max(b0, b1);
	a0 = a1 = 0; b0 = -0x3f3f3f3f; b1 = 0; //选择x
	for (int i = 1; i <= tot; i++) {
		a0 = max(b0, b1) + f[c[i]][0]; a1 = b0 + f[c[i]][1];
		b0 = a0; b1 = a1;
	}
	f[x][1] += b0;
}

void tarjan(int x) {
	dfn[x] = low[x] = ++tme;
	stk[++top] = x;
	f[x][1] = a[x];
	for (int i = head[x]; i; i = pre[i]) {
		int y = to[i];
		if (!dfn[y]) {
			tarjan(y);
			low[x] = min(low[x], low[y]);
			if (dfn[x] == low[y]) {
				int u = 0; tot = 0;
				do {
					u = stk[top--];
					c[++tot] = u;
				} while (u != y);
				solve(x);
			} else if (dfn[x] < low[y]) {
				f[x][0] += max(f[y][0], f[y][1]);
				f[x][1] += f[y][0];
			}
		} else low[x] = min(low[x], dfn[y]);
	}
}

int main() {
	scanf("%d %d", &n, &m);
	for (int i = 1, u, v; i <= m; i++) {
		scanf("%d %d", &u, &v);
		addedge(u, v);
	}
	for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
	tarjan(1);
	printf("%d\n", max(f[1][0], f[1][1]));
	return 0;
}