luogu P2763 试题库问题

本题可以用最大流也可以用最大匹配(本质一样)，用dinic最大流好建图，但码量大，匈牙利码量小，建图费点劲。

最大流：依旧是设一个源点一个汇点，对于每一个种类，连一条到汇点的边，capacity为需要的量，对于每一个试题，从源点连一条capacity为1的边到他，从他对每一个其所属的编号种类连一条capacity为1的边，求最大流即可，再找出最小割即可

#include<bits/stdc++.h>
using namespace std;
#define lowbit(x) ((x)&(-x))
typedef long long LL;

const int maxm = 1e5+5;
const int maxn = 1e3+33;
const int INF = 0x3f3f3f3f;

struct edge{
    int u, v, cap, flow, nex;
} edges[maxm];

int head[maxm], cur[maxm], cnt, level[maxn], capacity[23];
vector<int> ans[23];

void init() {
    memset(head, -1, sizeof(head));
}

void addedge(int u, int v, int cap) {
    edges[cnt] = edge{u, v, cap, 0, head[u]};
    head[u] = cnt++;
}

void bfs(int s) {
    memset(level, -1, sizeof(level));
    queue<int> q;
    level[s] = 0;
    q.push(s);
    while(!q.empty()) {
        int u = q.front();
        q.pop();
        for(int i = head[u]; i != -1; i = edges[i].nex) {
            edge& now = edges[i];
            if(now.cap > now.flow && level[now.v] < 0) {
                level[now.v] = level[u] + 1;
                q.push(now.v);
            }
        }
    }
}

int dfs(int u, int t, int f) {
    if(u == t) return f;
    for(int& i = cur[u]; i != -1; i = edges[i].nex) {
        edge& now = edges[i];
        if(now.cap > now.flow && level[u] < level[now.v]) {
            int d = dfs(now.v, t, min(f, now.cap - now.flow));
            if(d > 0) {
                now.flow += d;
                edges[i^1].flow -= d;
                return d;
            }

        }
    }
    return 0;
}

int dinic(int s, int t) {
    int maxflow = 0;
    for(;;) {
        bfs(s);
        if(level[t] < 0) break;
        memcpy(cur, head, sizeof(head));
        int f;
        while((f = dfs(s, t, INF)) > 0)
            maxflow += f;
    }
    return maxflow;
}

void run_case() {
    int k, n, p, u, sum = 0;
    init();
    cin >> k >> n;
    int s = 0, t = n+k+1;
    for(int i = 1; i <= k; ++i) {
        cin >> capacity[i];
        sum += capacity[i];
        addedge(n+i, t, capacity[i]), addedge(t, n+i, 0);
    }
    for(int i = 1; i <= n; ++i) {
        cin >> p;
        addedge(s, i, 1), addedge(i, s, 0);
        while(p--) {
            cin >> u;
            addedge(i, u+n, 1), addedge(u+n, i, 0);
        }
    }
    if(dinic(s, t) != sum) {cout << "No Solution!"; return;}
    for(int i = 1; i <= n; ++i) {
        for(int j = head[i]; j != -1; j = edges[j].nex) {
            if(edges[j].flow) {
                ans[edges[j].v - n].push_back(i);
                break;
            }
        }
    }
    for(int i = 1; i <= k; ++i) {
        cout << i << ":";
        for(auto j : ans[i])
            cout << " " << j;
        cout << "\n";
    }
}

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    run_case();
    //cout.flush();
    return 0;
}

Dinic

匈牙利：对于每一个种类都连一条边到其所属的试卷，求最大匹配即可(无代码)

原文地址：https://www.cnblogs.com/GRedComeT/p/12274919.html