poj 1698 二分图多重匹配

网络流解法：

容易想到将每一天看做一个点并和汇点连容量为1的边（因为一天只能做一件事情），每个电影看做一个点并和源点连容量为d的边（即几天可以完成），然后电影和对应天数连容量为1的边，求最大流判断是否满流即可。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int INF = 9999999;
const int N = 400;
const int M = 20000;
int head[N];
int level[N];
int q[N];
int n, e, front, rear;

void init()
{
    e = 0;
    memset( head, -1, sizeof(head) );
}

struct Edge 
{
    int v, next, cap;
} edge[M];

void addEdge( int u, int v, int cap )
{
    edge[e].v = v;
    edge[e].cap = cap;
    edge[e].next = head[u];
    head[u] = e++;
    edge[e].v = u;
    edge[e].cap = 0;
    edge[e].next = head[v];
    head[v] = e++;
}

bool bfs( int s, int t )
{
    front = rear = 0;
    memset( level, -1, sizeof(level) );
    q[rear++] = s;
    level[s] = 0;
    while ( front < rear )
    {
        int u = q[front++];
        for ( int i = head[u]; i != -1; i = edge[i].next )
        {
            int v = edge[i].v, cap = edge[i].cap;
            if ( level[v] == -1 && cap > 0 )
            {
                level[v] = level[u] + 1;
                q[rear++] = v;
            }
        }
    }
    return level[t] != -1;
}

int dfs( int u, int sum, int t )
{
    if ( u == t ) return sum;
    int os = 0;
    for ( int i = head[u]; i != -1; i = edge[i].next )
    {
        int v = edge[i].v, cap = edge[i].cap;
        if ( level[v] == level[u] + 1 && cap > 0 )
        {
            int tt = dfs( v, min( sum, cap ), t );
            if ( tt == 0 ) continue;
            edge[i].cap -= tt;
            edge[i ^ 1].cap += tt;
            sum -= tt;
            os += tt;
            if ( sum == 0 ) break;
        }
    }
    if ( os == 0 ) level[u] = -1;
    return os;
}

int dinic( int s, int t )
{
    int res = 0;
    while ( bfs( s, t ) )
    {
        res += dfs( s, INF, t );
    }
    return res;
}

const int D = 7;
int nn[D];

int main ()
{
    int _case;
    scanf("%d", &_case);
    while ( _case-- )
    {
        scanf("%d", &n);
        init();
        int tot = 0, wmax = -1, d, w;
        for ( int i = 1; i <= n; i++ )
        {
            for ( int j = 0; j < D; j++ )
            {
                scanf("%d", nn + j);
            }
            scanf("%d%d", &d, &w);
            tot += d;
            wmax = max( wmax, w );
            addEdge( 0, i, d );
            for ( int j = 0; j < D; j++ )
            {
                if ( nn[j] == 1 )
                {
                    for ( int k = 0; k < w; k++ )
                    {
                        int num = n + k * 7 + j + 1;
                        addEdge( i, num, 1 );
                    }
                }
            }
        }
        int tid = n + wmax * 7 + 1;
        for ( int i = n + 1; i < tid; i++ )
        {
            addEdge( i, tid, 1 );
        }
        if ( dinic( 0, tid ) == tot )
        {
            puts("Yes");
        }
        else
        {
            puts("No");
        }
    }
    return 0;
}

匈牙利解法（些许改变）：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
using namespace std;

const int X = 350;
const int Y = 20;
int head[X];
int sz[Y];
bool visit[Y];
vector<int> mark[Y];
int e;

void init()
{
    e = 0;
    memset( head, -1, sizeof(head) );
}

struct Edge 
{
    int v, next;
} edge[X * Y];

void addEdge( int u, int v )
{
    edge[e].v = v;
    edge[e].next = head[u];
    head[u] = e++;
}

int dfs( int u )
{
    for ( int i = head[u]; i != -1; i = edge[i].next )
    {
        int v = edge[i].v;
        if ( !visit[v] )
        {
            visit[v] = 1;
            if ( mark[v].size() < sz[v] )
            {
                mark[v].push_back(u);
                return 1;
            }
            else
            {
                for ( int j = 0; j < mark[v].size(); j++ )
                {
                    if ( dfs( mark[v][j] ) )
                    {
                        mark[v][j] = u;
                        return 1;
                    }
                }
            }
        }
    }
    return 0;
}

const int D = 7;
int nn[D];

int main ()
{
    int _case;
    scanf("%d", &_case);
    while ( _case-- )
    {
        int n;
        scanf("%d", &n);
        init();
        int tot = 0, wmax = -1, d, w;
        for ( int i = 0; i < n; i++ )
        {
            for ( int j = 0; j < D; j++ )
            {
                scanf("%d", nn + j);
            }
            scanf("%d%d", &d, &w);
            sz[i] = d;
            tot += d;
            wmax = max( wmax, w );
            for ( int j = 0; j < D; j++ )
            {
                if ( nn[j] == 1 )
                {
                    for ( int k = 0; k < w; k++ )
                    {
                        int num = k * 7 + j;
                        addEdge( num, i );
                    }
                }
            }
        }
        for ( int i = 0; i < n; i++ )
        {
            mark[i].clear();
        }
        int res = 0;
        for ( int i = 0; i < 7 * wmax; i++ )
        {
            memset( visit, 0, sizeof(visit) );
            res += dfs(i);
        }
        if ( res == tot )
        {
            puts("Yes");
        }
        else
        {
            puts("No");
        }
    }
    return 0;
}