洛谷P2763 试题库问题(最大流)

题意

$n$道试题，每道题有多种类别属性

抽取$m$道题组成试卷，要求包含指定的类型

输出方案

Sol

又是一道zz网络流

我的构图长这样，$k_i$表示第$i$道试题需要的数量

#include<cstdio>

#include<queue>

#include<cstring>

using namespace std;

const int MAXN = 1e5 + , INF = 1e9 + ;

inline int read() {

    char c = getchar(); int x = , f = ;

    while(c < '' || c > '') {if(c == '-') f = -; c = getchar();}

    while(c >= '' && c <= '') x = x *  + c - '', c = getchar();

    return x * f;

}

int K, N, S, T;

int need[MAXN];

vector<int>ans[MAXN];

struct Edge {

    int u, v, f, nxt;

}E[MAXN];

int head[MAXN], cur[MAXN], num;

inline void add_edge(int x, int y, int f) {

    E[num] = (Edge){x, y, f, head[x]};

    head[x] = num++;

}

inline void AddEdge(int x, int y, int z) {

    add_edge(x, y, z);

    add_edge(y, x, );

}

int sum = , deep[MAXN];

bool BFS() {

    queue<int> q; q.push(S);

    memset(deep, , sizeof(deep)); deep[S] = ;

    while(!q.empty()) {

        int p = q.front(); q.pop();

        for(int i = head[p]; i != -; i = E[i].nxt) {

            int to = E[i].v;

            if(!deep[to] && E[i].f) {

                deep[to] = deep[p] + ;

                q.push(to);

            }

        }

    }

    return deep[T] > ;

}

int DFS(int x, int flow) {

    if(x == T) return flow;

    int ansflow = ;

    for(int &i = cur[x]; i != -; i = E[i].nxt) {

        int to = E[i].v;

        if(deep[to] == deep[x] +  && E[i].f) {

            int nowflow = DFS(to, min(flow, E[i].f));

            E[i].f -= nowflow; E[i ^ ].f += nowflow;

            ansflow += nowflow; flow -= nowflow;

            if(flow <= ) break;

        }

    }

    return ansflow;

}

int Dinic() {

    int ans = ;

    while(BFS()) {

        memcpy(cur, head, sizeof(head));

        ans += DFS(S, INF);

    }

    return ans;

}

int main() {

    memset(head, -, sizeof(head));

    K = read(); N = read(); S = ; T = N + K + ;

    int sum = ;

    for(int i = ; i <= K; i++) {

        int x = read(); sum += x;

        AddEdge(N + i, T, x);

    }

    for(int i = ; i <= N; i++) {

        int n = read();

        AddEdge(S, i, );

        for(int j = ; j <= n; j++) {

            AddEdge(i, read() + N, );

        }

    }

    if(Dinic() < sum) {puts("No Solution!"); return ;}

    for(int x = ; x <= N; x++)

        for(int i = head[x]; i != -; i = E[i].nxt)

            if(E[i].f == )

                ans[E[i].v - N].push_back(x);

    for(int i = ; i <= K; i++) {

        printf("%d: ", i);

        for(int j = ; j < ans[i].size(); j++) printf("%d ", ans[i][j]);

        puts("");

    }

    return ;

}