CF 976F 递增容量最大流

 

给你一个二分图 要求你求出对于k=[0~Mindegree] 每个点的度数至少为k所需要的最少边数 并输出方案

如果是单个询问的话 直接跑一个下界网络流即可 但是有多个询问 重建图强行跑不行

反过来考虑,变成至多能删除多少边则建边[s,i,degree[i]-Mindegree] [i,T,degree[i]-Mindegree] [u,v,1]

这样跑出来的流 二分图中没有流量的边代表是要选的 有流量的是要删的 同时保证了每个点的度数不小于Mindegree

则接下来每次对与S,T相连的边容量++ 得到k=[0~Mindegree-1]的答案

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 4050;
const int MAXM = 10005;
const int INF = 1000000050;
int Head[MAXN], cur[MAXN], lev[MAXN], to[MAXM << 1], nxt[MAXM << 1], f[MAXM << 1], ed = 1, S, T;
inline void addedge(int u, int v, int cap)
{
    to[++ed] = v;
    nxt[ed] = Head[u];
    Head[u] = ed;
    f[ed] = cap;
    to[++ed] = u;
    nxt[ed] = Head[v];
    Head[v] = ed;
    f[ed] = 0;
    return;
}
inline bool BFS()
{
    int u;
    memset(lev, -1, sizeof(lev));
    queue<int>q;
    lev[S] = 0;
    q.push(S);
    while (q.size()) {
        u = q.front();
        q.pop();
        for (int i = Head[u]; i; i = nxt[i])
            if (f[i] && lev[to[i]] == -1) {
                lev[to[i]] = lev[u] + 1;
                q.push(to[i]);
                /*
                if (to[i] == T)
                {
                        return 1;
                }
                magic one way optimize
                */
            }
    }
    memcpy(cur, Head, sizeof Head);
    return lev[T] != -1;
}
inline int DFS(int u, int maxf)
{
    if (u == T || !maxf) {
        return maxf;
    }
    int cnt = 0;
    for (int &i = cur[u], tem; i; i = nxt[i])
        if (f[i] && lev[to[i]] == lev[u] + 1) {
            tem = DFS(to[i], min(maxf, f[i]));
            maxf -= tem;
            f[i] -= tem;
            f[i ^ 1] += tem;
            cnt += tem;
            if (!maxf) {
                break;
            }
        }
    if (!cnt) {
        lev[u] = -1;
    }
    return cnt;
}
int Dinic()
{
    int ans = 0;
    while (BFS()) {
        ans += DFS(S, 2147483647);
    }
    return ans;
}
void init(int SS, int TT)
{
    memset(Head, 0, sizeof(Head));
    ed = 1;
    S = SS;
    T = TT;
    return;
}
int du[4005];
int ans[4005][4005];
int main()
{
    int n1, n2, m;
    int u, v;
    scanf("%d %d %d", &n1, &n2, &m);
    int n = n1 + n2;
    for (int i = 1; i <= m; i++) {
        scanf("%d %d", &u, &v);
        addedge(u, v + n1, 1);
        du[u]++, du[v + n1]++;
    }
    int Mindegree = INT_MAX;
    for (int i = 1; i <= n; i++) {
        Mindegree = min(Mindegree, du[i]);
    }
    S = 0, T = n + 1;
    for (int i = 1; i <= n1; i++) {
        addedge(S, i, du[i] - Mindegree);
    }
    for (int i = n1 + 1; i <= n; i++) {
        addedge(i, T, du[i] - Mindegree);
    }
    int ansnow = Dinic();
    for (int x = 1; x <= n1; x++) {
        for (int i = Head[x]; i; i = nxt[i]) {
            v = to[i];
            if (v >= n1 + 1 && v <= n) {
                if (f[i] == 1) {
                    ans[Mindegree][++ans[Mindegree][0]] = i / 2;
                }
            }
        }
    }
    for (int i = Head[S]; i; i = nxt[i]) {
        f[i]++;
    }
    for (int x = n1 + 1; x <= n; x++) {
        for (int i = Head[x]; i; i = nxt[i]) {
            v = to[i];
            if (v == T) {
                f[i]++;
            }
        }
    }
    for (int i = Mindegree - 1; i >= 0; i--) {
        ansnow = Dinic();
        for (int x = 1; x <= n1; x++) {
            for (int j = Head[x]; j; j = nxt[j]) {
                v = to[j];
                if (v >= n1 + 1 && v <= n) {
                    if (f[j] == 1) {
                        ans[i][++ans[i][0]] = j / 2;
                    }
                }
            }
        }
        for (int j = Head[S]; j; j = nxt[j]) {
            f[j]++;
        }
        for (int x = n1 + 1; x <= n; x++) {
            for (int j = Head[x]; j; j = nxt[j]) {
                v = to[j];
                if (v == T) {
                    f[j]++;
                }
            }
        }
    }
    for (int i = 0; i <= Mindegree; i++) {
        printf("%d", ans[i][0]);
        for (int j = 1; j <= ans[i][0]; j++) {
            printf(" %d", ans[i][j]);
        }
        puts("");
    }
    return 0;
}