[Codeforces Gym - 101480J] Juice Junctions

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/u013578420/article/details/81915440

题目大意：
给你一个n个点m条边的无向图， 求

$$\sum_{s}\sum_{s<t} s到t的最大流$$
保证每个点的度数 $\leq 3$。 $(n \leq 3000, m \leq 4500)$

题目思路：
由于每个点的度数$\leq 3$， 由最大流最小割定理， 最大流只可能有{0, 1, 2, 3}4种情况。
对于$\leq 1$的答案， 点对在一个连通块内。
对于$\leq 2$的答案， 点对在一个双联通内。
对于$\leq 3$的答案， 枚举删除任意一条边， 点对一直在同一个双联通内(可以用hash判断)。

#include <map>
#include <cmath>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

#define ll long long
#define ull unsigned ll
#define db double
#define fi first
#define se second
#define pi pair<int, int >
#define ls (x << 1)
#define rs ((x << 1) | 1)
#define mid ((l + r) >> 1)
#define mp(x, y) make_pair((x), (y))
#define pb push_back

using namespace std;

const int N = 3010;
const int M = 5010;

int n, m;
int cnt = 1, lst[N], nxt[M * 2], to[M * 2];

void add(int u, int v){
    nxt[++ cnt] = lst[u]; to[cnt] = v; lst[u] = cnt;
    nxt[++ cnt] = lst[v]; to[cnt] = u; lst[v] = cnt;
}

int ans;
int indx, dfn[N], low[N]; ull id[N];
int top, stk[N], instk[N]; bool ban[M];

void tarjan(int u, int pre){
    dfn[u] = low[u] = ++ indx;
    stk[++ top] = u; instk[u] = 1;
    for (int j = lst[u]; j; j = nxt[j]){
        int v = to[j];
        if (j == (pre ^ 1)) continue;
        if (ban[j >> 1]) continue;
        if (!dfn[v]){
            tarjan(v, j);
            low[u] = min(low[u], low[v]);
        }
        else if (instk[v]) low[u] = min(low[u], dfn[v]);
    }
    if (low[u] == dfn[u]){
        while (stk[top] != u){
            id[stk[top]] = id[stk[top]] * 233 + u;
            instk[stk[top]] = 0;
            top --;
        }
        id[u] = id[u] * 233 + u;
        instk[u] = 0;
        top --;
    }
}

int main(){
    scanf("%d %d", &n, &m);
    for (int i = 1, u, v; i <= m; i ++){
        scanf("%d %d", &u, &v); add(u, v);
    }

    for (int i = 1; i <= n; i ++)
        if (!dfn[i]){
            int preindx = indx;
            tarjan(i, 0);
            ans += (indx - preindx) * (indx - preindx - 1) / 2;
        }

    for (int i = 1; i <= n; i ++)
        for (int j = i + 1; j <= n; j ++)
            if (id[i] == id[j]) ans ++;

    for (int j = 1; j <= m; j ++){
        indx = 0;
        memset(dfn, 0, sizeof(dfn));

        ban[j] = 1;
        for (int i = 1; i <= n; i ++)
            if (!dfn[i]) tarjan(i, 0);

        ban[j] = 0;
    }

    for (int a = 1; a <= n; a ++)
        for (int b = a + 1; b <= n; b ++){
            if (id[a] == id[b]) ans ++;
        }

    printf("%d\n", ans);
    return 0;
}