HDU 2242 考研路茫茫——空调教室

题目链接

思路：求边双连通分量，然后进行缩点，点权为双连通分支的点权之和，缩点完变成一棵树，然后在树上dfs一遍就能得出答案

代码：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <vector>
using namespace std;

const int N = 10005;
const int M = 20005;

int n, m, val[N];

struct Edge {
	int u, v, id;
	bool iscut;
	Edge() {}
	Edge(int u, int v, int id) {
		this->u = u;
		this->v = v;
		this->id = id;
		this->iscut = false;
	}
} edge[M * 2], cut[M];

int en, first[N], next[M], cutn;

void add_edge(int u, int v, int id) {
	edge[en] = Edge(u, v, id);
	next[en] = first[u];
	first[u] = en++;
}

int pre[N], dfn[N], bccno[N], bccval[N], bccn, dfs_clock;

void dfs_cut(int u, int fa) {
	pre[u] = dfn[u] = ++dfs_clock;
	for (int i = first[u]; i + 1; i = next[i]) {
		int v = edge[i].v;
		if (edge[i].id == fa) continue;
		if (!pre[v]) {
			dfs_cut(v, edge[i].id);
			dfn[u] = min(dfn[u], dfn[v]);
			if (dfn[v] > pre[u]) {
				edge[i].iscut = edge[i^1].iscut = true;
				cut[cutn++] = edge[i];
			}
		} else dfn[u] = min(dfn[u], pre[v]);
	}
}

void find_cut() {
	dfs_clock = 0; cutn = 0;
	memset(pre, 0, sizeof(pre));
	for (int i = 0; i < n; i++)
		if (!pre[i]) dfs_cut(i, -1);
}

void dfs_bcc(int u) {
	pre[u] = 1;
	bccno[u] = bccn;
	bccval[bccn] += val[u];
	for (int i = first[u]; i + 1; i = next[i]) {
		if (edge[i].iscut) continue;
		int v = edge[i].v;
		if (pre[v]) continue;
		dfs_bcc(v);
	}
}

vector<int> bcc[N];

void find_bcc() {
	bccn = 0;
	memset(bccval, 0, sizeof(bccval));
	memset(pre, 0, sizeof(pre));
	for (int i = 0; i < n; i++) {
		if (!pre[i]) {
			dfs_bcc(i);
			bccn++;
		}
	}
}

const int INF = 0x3f3f3f3f;

int ans, tot;

int gao(int u, int fa) {
	int sum = bccval[u];
	for (int i = 0; i < bcc[u].size(); i++) {
		int v = bcc[u][i];
		if (v == fa) continue;
		int tmp = gao(v, u);
		sum += tmp;
		ans = min(ans, abs(tot - 2 * tmp));
	}
	return sum;
}

int main() {
	while (~scanf("%d%d", &n, &m)) {
		en = 0;
		memset(first, -1, sizeof(first));
		tot = 0;
		for (int i = 0; i < n; i++) {
			scanf("%d", &val[i]);
			tot += val[i];
		}
		int u, v;
		for (int i = 0; i < m; i++) {
			scanf("%d%d", &u, &v);
			add_edge(u, v, i);
			add_edge(v, u, i);
		}
		find_cut();
		find_bcc();
		if (cutn == 0) {
			printf("impossible\n");
			continue;
		}
		for (int i = 0; i < bccn; i++) bcc[i].clear();
		for (int i = 0; i < cutn; i++) {
			int u = bccno[cut[i].u];
			int v = bccno[cut[i].v];
			bcc[u].push_back(v);
			bcc[v].push_back(u);
		}
		ans = INF;
		gao(0, -1);
		printf("%d\n", ans);
	}
	return 0;
}