HDU 2242 考研路茫茫——空调教室

大意：给定一个无向图，问是否存在一条边使得删去该边后，使得该边左、右的双联通分量的权值的差值尽量小。

思路：求得双连通分量后，把原图缩成了一棵树，然后再树上做DP即可，求ans = min(ans, (sum-val*2)); 其中val为左右子树的权值。

 

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <map>
using namespace std;

const int maxn = 10010;
const int maxm = 200010*2;
const int INF = 0x3f3f3f3f;

struct Edge
{
    int u, v, w;
    int next;
    int id;
}edge[maxm], edge2[maxm];

int first[maxn], stack[maxn], ins[maxn], dfn[maxn], low[maxn];
int first2[maxn];
int belong[maxn];
int ind[maxn], outd[maxn];

int cost[maxn];
int w[maxn];

int n, m;
int cnt, cnt2;
int scnt, top, tot;

void init()
{
    cnt = cnt2 = 0;
    scnt = top = tot = 0;
    memset(first, -1, sizeof(first));
    memset(first2, -1, sizeof(first2));
    memset(dfn, 0, sizeof(dfn));
    memset(ins, 0, sizeof(ins));
    memset(ind, 0, sizeof(ind));
    memset(cost, 0, sizeof(cost));
}

void read_graph(int u, int v)
{  
    edge[cnt].u = u, edge[cnt].v = v;
    edge[cnt].next = first[u], first[u] = cnt++;  
}

void read_graph2(int u, int v, int w)  
{  
    edge2[cnt2].u = u, edge2[cnt2].v = v, edge2[cnt2].w = w;
    edge2[cnt2].next = first2[u], first2[u] = cnt2++;  
}  

vector<Edge> bridge;

void dfs(int u, int fa)
{
    int v;
    low[u] = dfn[u] = ++tot;
    stack[top++] = u;
    ins[u] = 1;
    bool repeat = 0;
    for(int e = first[u]; e != -1; e = edge[e].next)
    {
        v = edge[e].v;
        if(v == fa && !repeat) { repeat = 1; continue; }
        if(!dfn[v])
        {
            dfs(v, u);
            low[u] = min(low[u], low[v]);
            if(dfn[u] < low[v])
            {
                bridge.push_back(edge[e]); //保存割边 
            }
        }
        else if(ins[v]) { low[u] = min(low[u], dfn[v]); }
    }
    if(low[u] == dfn[u])
    {
        scnt++;
        do { v = stack[--top]; belong[v] = scnt; ins[v] = 0; cost[scnt] += w[v]; } while(u != v);
    }
}

void Tarjan()
{
    for(int v = 1; v <= n; v++) if(!dfn[v])
        dfs(v, -1);
}

int sum;

void read_case()
{
    init();
    sum = 0;
    for(int i = 1; i <= n; i++)
    {
        scanf("%d", &w[i]);
        sum += w[i];
    }
    while(m--)
    {
        int u, v;
        scanf("%d%d", &u, &v); u++; v++;
        read_graph(u, v), read_graph(v, u);
    }
}

void build()
{
    Tarjan();
    for(int u = 1; u <= n; u++)
    {
        for(int e = first[u]; e != -1; e = edge[e].next)
        {
            int v = edge[e].v;
            if(belong[u] != belong[v])
            {
                read_graph2(belong[u], belong[v], cost[belong[v]]);
            }
        }
    }
}

int d[maxn];
bool vis[maxn];

int dp(int u, int fa, int &ans)
{
    int val = cost[u];
    for(int e = first2[u]; e != -1; e = edge2[e].next)
    {
        int v = edge2[e].v;
        if(v == fa) continue; //重要 
        val += dp(v, u, ans);
    }
    ans = min(ans, abs(sum-val*2));
    return val;
}

void solve()
{
    read_case();
    build();
    if(scnt == 1) { printf("impossible\n"); return; }
    int ans = INF;
    dp(1, -1, ans);
    printf("%d\n", ans);
}

int main()
{
    while(~scanf("%d%d", &n, &m))
    {
        solve();
    }
    return 0;
}