HDU 2242 考研路茫茫——空调教室(边双连通）

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
;

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
, 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
, dfn
, bccno
, bccval
, 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
;

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;
}