HDU 2242 双连通分量 考研路茫茫——空调教室

思路就是求边双连通分量，然后缩点，再用树形DP搞一下。

代码和求强连通很类似，有点神奇，=_=，慢慢消化吧

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

const int maxn = 10000 + 10;
const int maxm = 20000 + 10;

int n, m;
int a[maxn];

struct Edge
{
    int v, nxt;
}edges[maxm * 2];
int ecnt;
int head[maxn];

bool vis[maxn];

void AddEdge(int u, int v)
{
    edges[ecnt].v = v;
    edges[ecnt].nxt = head[u];
    head[u] = ecnt++;
}

stack<int> S;
int dfs_clock, scc_cnt;
int low[maxn], pre[maxn], sccno[maxn], w[maxn];

void dfs(int u, int fa)
{
    low[u] = pre[u] = ++dfs_clock;
    S.push(u);

    for(int i = head[u]; ~i; i = edges[i].nxt)
    {
        if(i == (fa ^ 1)) continue;
        int v = edges[i].v;
        if(!pre[v])
        {
            dfs(v, i);
            low[u] = min(low[u], low[v]);
        }
        else if(!sccno[v]) low[u] = min(low[u], pre[v]);
    }

    if(pre[u] == low[u])
    {
        scc_cnt++;
        for(;;)
        {
            int x = S.top(); S.pop();
            sccno[x] = scc_cnt;
            w[scc_cnt] += a[x];
            if(x == u) break;
        }
    }
}

void find_scc()
{
    scc_cnt = dfs_clock = 0;
    memset(w, 0, sizeof(w));
    memset(pre, 0, sizeof(pre));
    memset(sccno, 0, sizeof(sccno));
    for(int i = 0; i < n; i++) if(!pre[i]) dfs(i, -1);
}

vector<int> G[maxn];

void dfs2(int u)
{
    vis[u] = true;
    for(int i = 0; i < G[u].size(); i++)
    {
        int v = G[u][i];
        if(vis[v]) continue;
        dfs2(v);
        w[u] += w[v];
    }
}

int main()
{
    while(scanf("%d%d", &n, &m) == 2 && n)
    {
        int sum = 0;
        for(int i = 0; i < n; i++) { scanf("%d", a + i); sum += a[i]; }

        memset(head, -1, sizeof(head));
        ecnt = 0;

        while(m--)
        {
            int u, v; scanf("%d%d", &u, &v);
            AddEdge(u, v); AddEdge(v, u);
        }

        find_scc();

        if(scc_cnt == 1) { puts("impossible"); continue; }

        for(int i = 1; i <= scc_cnt; i++) G[i].clear();
        for(int u = 0; u < n; u++)
            for(int i = head[u]; ~i; i = edges[i].nxt)
            {
                int v = edges[i].v;
                if(sccno[u] != sccno[v]) G[sccno[u]].push_back(sccno[v]);
            }

        memset(vis, false, sizeof(vis));
        dfs2(1);

        int ans = 1000000000;
        for(int i = 1; i <= scc_cnt; i++) ans = min(ans, abs(sum - w[i] * 2));
        printf("%d
", ans);
    }

    return 0;
}