【最大流-Dinic算法】B - Dining POJ - 3281

偏执的太偏执、 2022-06-10 08:41 179阅读 0赞

Think:  
1知识点：【最大流-Dinic算法】+【建图】  
2题意：输入N, F, D, 代表N头奶牛，F个食物，D个饮料，每头奶牛只吃喜欢的食物，只喝喜欢的饮料，询问最多有多少头奶牛可以满足食物和饮料各一个  
3思路：

/*
    1:[1, F]:食物结点编号
    2:[F+1, F+N*2]:奶牛结点编号
    3:[F+N*2+1, F+N*2+D]:饮料结点编号
    4:将奶牛拆点
      u = F+i, v = F+N+i, w = 1;
    */

以下为Accepted代码

#include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <queue>
    
    using namespace std;
    
    const int inf = 0x3f3f3f3f;
    const int M = 81400;
    
    struct Edge{
        int v;
        int cap;
        int next;
    }edge[M<<1];
    
    int cnt, head[414], level[414];
    
    void Init();
    void add_edge(int u, int v, int w);
    bool bfs(int s, int t);
    int dfs(int s, int t, int f);
    int max_flow(int s, int t);
    
    int main(){
        int N, F, D, i, Fi, Di, x, s, t;
        while(~scanf("%d %d %d", &N, &F, &D)){
            Init();
            s = 0, t = F + N*2 + D + 1;
    
            for(i = 1; i <= F; i++)
                add_edge(s, i, 1);
    
            for(i = 1; i <= N; i++){
                scanf("%d %d", &Fi, &Di);
                while(Fi--){
                    scanf("%d", &x);
                    add_edge(x, F+i, 1);
                }
                add_edge(F+i, F+N+i, 1);
                while(Di--){
                    scanf("%d", &x);
                    add_edge(F+N+i, F+N*2+x, 1);
                }
            }
    
            for(i = 1; i <= D; i++)
                add_edge(F+N*2+i, t, 1);
    
            printf("%d\n", max_flow(s, t));
        }
    }
    void Init(){
        cnt = 0;
        memset(head, -1, sizeof(head));
    }
    void add_edge(int u, int v, int w){
        edge[cnt].v = v;
        edge[cnt].cap = w;
        edge[cnt].next = head[u];
        head[u] = cnt++;
    
        edge[cnt].v = u;
        edge[cnt].cap = 0;
        edge[cnt].next = head[v];
        head[v] = cnt++;
    }
    bool bfs(int s, int t){
        queue <int> q;
        memset(level, -1, sizeof(level));
        level[s] = 0;
        q.push(s);
        while(!q.empty()){
            int u = q.front();
            q.pop();
            for(int i = head[u]; ~i; i = edge[i].next){
                int v = edge[i].v;
                if(edge[i].cap > 0 && level[v] == -1){
                    level[v] = level[u] + 1;
                    q.push(v);
                }
            }
        }
        if(level[t] == -1)
            return false;
        else
            return true;
    }
    int dfs(int u, int t, int f){
        if(u == t)
            return f;
        int flow = 0, d;
        for(int i = head[u]; ~i; i = edge[i].next){
            int v = edge[i].v;
            if(edge[i].cap > 0 && level[u] < level[v] && (d = dfs(v, t, min(edge[i].cap, f)))){
                edge[i].cap -= d;
                edge[i^1].cap += d;
                flow += d;
                f -= d;
                if(!f)
                    break;
            }
        }
        if(flow == 0)
            level[u] = -1;
        return flow;
    }
    int max_flow(int s, int t){
        int flow = 0;
        while(bfs(s, t)){
            flow += dfs(s, t, inf);
        }
        return flow;
    }