hdu2242 （边双缩点）

题意：给出n个点和m条无向边，每个点都有权值，要求去掉一条边，使得整个图变成两个连通块，并且两者的权值总和之差最小
**思路：**首先去掉的边必定是桥，无桥的话对应无解，有桥的话，把边双都求出来缩点，然后dfs一遍统计答案

#include<bits/stdc++.h>
using namespace std;
#define ls rt<<1
#define rs (rt<<1)+1
#define PI acos(-1)
#define eps 1e-8
#define ll long long
#define fuck(x) cout<<#x<<"     "<<x<<endl;
typedef pair<int,int> pii;
const int inf=2e9;
const int maxn=1e4+10;
int d[4][2]={1,0,-1,0,0,1,0,-1};
//int lowbit(int x){return x&-x;}
//void add(int x,int v){while(x<=n)bit[x]+=v,x+=lowbit(x);}
//int sum(int x){int ans=0;while(x>=1) ans+=bit[x],x-=lowbit(x);return ans;}
inline ll read() {
    ll s = 0,w = 1;
    char ch = getchar();
    while(!isdigit(ch)) {
        if(ch == '-') w = -1;
        ch = getchar();
    }
    while(isdigit(ch))
        s = s * 10 + ch - '0',ch = getchar();
    return s * w;
}
inline void write(ll x) {
    if(x < 0)
        putchar('-'), x = -x;
    if(x > 9)
        write(x / 10);
    putchar(x % 10 + '0');
}
int gcd(int x,int y){
    return y==0?x:gcd(y,x%y);
}


struct edge{
    int u,v,nxt;
}e[maxn<<2];
int sz[maxn],bccv[maxn],val[maxn],head[maxn],dfn[maxn],low[maxn],bcc[maxn],bccn,cnt,n,m,sum,minn;
stack<int>sta;
vector<int>g[maxn];

void dfs(int now,int fa){
    bool flag=0;
    dfn[now]=low[now]=++cnt;
    sta.push(now);
    for(int i=head[now];i!=-1;i=e[i].nxt){
        int v=e[i].v;
        if(!dfn[v]){
            dfs(v,now);
            low[now]=min(low[now],low[v]);
        }
        else
            if(v!=fa||flag)
                low[now]=min(low[now],dfn[v]);
            else
                flag=1;
    }
    if(low[now]==dfn[now]){
        bccn++;
        for(;;){
            int tmp=sta.top();sta.pop();
            bcc[tmp]=bccn;
            bccv[bccn]+=val[tmp];
            if(tmp==now) break;
        }
    }
}

void dfss(int now,int fa){
    sz[now]=bccv[now];
    for(int i=0;i<g[now].size();i++)
    {
        int v=g[now][i];
        if(v==fa||sz[v]!=0) continue;
        dfss(v,now);
        sz[now]+=sz[v];
        minn=min(minn,abs(sum-sz[v]-sz[v]));
    }
}

int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        int tmp=0;
        sum=cnt=bccn=0;
        minn=2e9;
        for(int i=1;i<=n;i++)
            head[i]=-1,dfn[i]=0,low[i]=0,sz[i]=0,bccv[i]=0,g[i].clear();
        for(int i=1;i<=n;i++)
            scanf("%d",&(val[i])),sum+=val[i];
        for(int i=1;i<=m;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            u++;v++;
            e[++tmp]={u,v,head[u]};
            head[u]=tmp;
            e[++tmp]={v,u,head[v]};
            head[v]=tmp;
        }
        for(int i=1;i<=n;i++)
            if(!dfn[i])
                dfs(i,-1);
        for(int i=1;i<=tmp;i+=2){
            int u=e[i].u,v=e[i].v;
            if(bcc[u]!=bcc[v])
                g[bcc[u]].push_back(bcc[v]),g[bcc[v]].push_back(bcc[u]);
        }
        dfss(1,-1);
        if(bccn==1){
            puts("impossible");
            continue;
        }
        write(minn);
        puts("");
    }
    return 0;
}