hiho1393二分图多重匹配

Posted 2020-09-07 Pealicx

题目链接：【http://hihocoder.com/problemset/problem/1393】

题意：中文题意。

题解：二分图的多重匹配。主要是建图然后跑一个最带流，再判断一下就可以了。

建图：首先要保证每个学生最多选择a[i]节课，那么我们建立一个超级起点S,S->学生，流量为学生最多选的课数，然后每个学生向它喜欢的课程建立一条流量为1的边，然后又要保证每个项目的人数，每个项目向超级汇点T建立一条流量为M[i]的边，表示每节课最多上M[i]个人。然后跑最大流，判断指向终点的边是不是满载就可以了。

【http://blog.csdn.net/tramp_1/article/details/52663763】推荐博客。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 1050;
const int maxm = 10050;
const int INF = 0x3f3f3f3f;
struct Edge
{
    int to, next, cap, flow;
    Edge(int to = 0, int next = 0, int cap = 0, int flow = 0): to(to), next(next), cap(cap), flow(flow) {}
} edge[maxm];
int head[maxn], tot;
void init()
{
    memset(head, -1, sizeof(head));
    tot = 0;
}
void addedge(int u, int v, int w, int rw = 0)
{
    edge[tot] = Edge(v, head[u], w, 0);
    head[u] = tot++;
    edge[tot] = Edge(u, head[v], rw, 0);
    head[v] = tot++;
}
int gap[maxn], dep[maxn], pre[maxn], cur[maxn];
int sap(int st, int ed, int nodenum)
{
    memset(gap, 0, sizeof(gap));
    memset(dep, 0, sizeof(dep));
    memcpy(cur, head, sizeof(head));
    int u = st;
    pre[u] = -1;
    gap[0] = nodenum;
    int ans = 0;
    while(dep[st] < nodenum)
    {
        if(u == ed)
        {
            int Min = INF;
            for(int i = pre[u]; i != -1; i = pre[edge[i ^ 1].to])
                if(Min > edge[i].cap - edge[i].flow)
                    Min = edge[i].cap - edge[i].flow;
            for(int i = pre[u]; i != -1; i = pre[edge[i ^ 1].to])
            {
                edge[i].flow += Min;
                edge[i ^ 1].flow -= Min;
            }
            u = st;
            ans += Min;
            continue;
        }
        bool fg = false;
        int v;
        for(int i = cur[u]; i != -1; i = edge[i].next)
        {
            v = edge[i].to;
            if(edge[i].cap - edge[i].flow && dep[v] + 1 == dep[u])
            {
                fg = true;
                cur[u] = pre[v] = i;
                break;
            }
        }
        if(fg)
        {
            u = v;
            continue;
        }
        int Min = nodenum;
        for(int i = head[u]; i != -1; i = edge[i].next)
            if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
            {
                Min = dep[edge[i].to];
                cur[u] = i;
            }
        gap[dep[u]]--;
        if(!gap[dep[u]]) return ans;
        dep[u] = Min + 1;
        gap[dep[u]]++;
        if(u != st) u = edge[pre[u] ^ 1].to;
    }
    return ans;
}
int T, n, m;
int main ()
{
    scanf("%d", &T);
    while(T--)
    {
        init();
        scanf("%d%d", &n, &m);
        int num = 0;
        for(int i = 1; i <= m; i++)
        {
            int t;
            scanf("%d", &t);
            num += t;
            addedge(n + i, n + m + 1, t);
        }
        for(int i = 1; i <= n; i++)
        {
            int x, t;
            scanf("%d%d", &t, &x);
            addedge(0, i, t);
            for(int j = 1; j <= x; j++)
            {
                scanf("%d", &t);
                addedge(i, n + t, 1);
            }
        }
        int flow = sap(0, n + m + 1, n + m + 2);
        if(flow == num)
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}