HDU 2242 （Tarjan双连通缩点+树形DP）

n个点m条无向边，要去一条边得到两个连通分支，所以要去掉的一定是桥。先跑一边Tarjan找到所有的桥，然后双连通缩点，所有的双连通分量所成一个点，再用桥连接，得到一棵树。然后跑一遍树形DP。

        代码：

// Header.#include <algorithm>#include <iostream>#include <sstream>#include <cstring>#include <cstdio>#include <vector>#include <string>#include <bitset>#include <queue>#include <cmath>#include <ctime>#include <set>#include <map>using namespace std;// Macrotypedef long long LL;#define TIME cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s." << endl;#define IN freopen("/Users/apple/input.txt", "r", stdin);#define OUT freopen("/Users/apple/out.txt", "w", stdout);#define mem(a, n) memset(a, n, sizeof(a))#define rep(i, n) for(int i = 0; i < (n); i ++)#define repD(i, n) for(int i = (n); i; i --)#define REP(i, t, n) for(int i = (t); i < (n); i ++)#define REPD(i, t, n) for(int i = (n); i > (t); i --)#define FOR(i, t, n) for(int i = (t); i <= (n); i ++)#define FORD(i, t, n) for(int i = (n); i >= (t); i --)#define ALL(v) v.begin(), v.end()#define Min(a, b) a = min(a, b)#define Max(a, b) a = max(a, b)#define put(a) printf("%d\n", a)#define ss(a) scanf("%s", a)#define si(a) scanf("%d", &a)#define sii(a, b) scanf("%d%d", &a, &b)#define siii(a, b, c) scanf("%d%d%d", &a, &b, &c)#define VI vector<int>#define pb push_back#define x first#define y secondconst int inf = 0x3f3f3f3f, N = 1e4 + 5, MOD = 1e9 + 7;// Macro endint T, cas = 0;int n, m, ne, ans, all, top, bch, dfsNum;int a[N], head[N], dfn[N], low[N], tree[N], sum[N], col[N], stack[N];vector<pair<int, int> > brige;bool vis[N];struct edge {    int v, next;}e[10 * N];// Impvoid addEdge(int u, int v) {    e[ne].v = v, e[ne].next = head[u], head[u] = ne ++;}void Tarjan(int u, int fa) {    dfn[u] = low[u] = dfsNum ++;    vis[u] = 1;    stack[top++] = u;    int flag = 0;    for(int i = head[u]; i != -1; i = e[i].next) {        int v = e[i].v;        if(v == fa && !flag) { flag = 1; continue; }        if(!vis[v]) {            Tarjan(v, u);            Min(low[u], low[v]);        } else if(col[v] == -1) Min(low[u], dfn[v]);        if(dfn[u] < low[v]) brige.pb(make_pair(u, v));    }    if(low[u] == dfn[u]) {        while(stack[top-1] != u) col[stack[--top]] = bch, sum[bch] += a[stack[top]];        top --;        sum[bch] += a[stack[top]];        col[u] = bch ++;    }}void build() {    int sz = brige.size();    rep(i, sz) {        int u = brige[i].x, v = brige[i].y;        e[ne].v = col[v], e[ne].next = tree[col[u]];        tree[col[u]] = ne ++;        e[ne].v = col[u], e[ne].next = tree[col[v]];        tree[col[v]] = ne ++;    }}int dfs(int u) {    int tmp = sum[u];    vis[u] = 1;    for(int i = tree[u]; i != -1; i = e[i].next) {        if(!vis[e[i].v]) tmp += dfs(e[i].v);    }    Min(ans, abs(all - 2 * tmp));    return tmp;}void init() {    mem(tree, -1);    mem(head, -1);    mem(col, -1);    mem(vis, 0);    mem(sum, 0);    brige.clear();    bch = top = all = ne = dfsNum = 0;}int main(){#ifdef LOCAL    IN // OUT#endif        while(sii(n, m) != EOF) {        init();        rep(i, n) si(a[i]), all += a[i];        int u, v;        rep(i, m) {            sii(u, v);            addEdge(u, v), addEdge(v, u);        }        Tarjan(0, 0);        if(bch != 1) {            build();            mem(vis, 0);            ans = inf;            dfs(0);            put(ans);        } else puts("impossible");    }        return 0;}