POJ 1698 Alice's Chance

/* 一个演员很想参加film，但是一天只能参加一项，且每项必须在规定的几个星期内参加规定的天数，求演员是否能全部完成所有的film档期。 可以考虑用网络流做，加源汇点，如果film_i 和week_i_day_j天有联系，那么就连一条INF边，源点到film_i连一条day_i边，各个星期的天数到汇点连一条1边，求最大流是否和总天数相等。最大流用isap就可以了，非常快。 */

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <memory.h>
#include <queue>
using namespace std;

#define INF 0x7ffffff
#define MAXN 500
#define MAXE 20000

struct Edge{
    int next,pair;
    int v,cap,flow;
}edge[MAXE];
int net[MAXN];
int nv,ne,s,t,index,max_week,sum;
void init()
{
    index = 0;
    memset(net,-1,sizeof(net));
    s = 0; // start position
    max_week = -1;
    sum = 0;
}
void add_edge(const int& u,const int& v,const int& val)
{
    edge[index].next = net[u];
    net[u] = index;
    edge[index].v = v;
    edge[index].cap = val;
    edge[index].flow = 0;
    edge[index].pair = index + 1;
    ++index;
    edge[index].next = net[v];
    net[v] = index;
    edge[index].v = u;
    edge[index].cap = 0;
    edge[index].flow = 0;
    edge[index].pair = index - 1;
    ++index;
}
int ISAP()
{
    int numb[MAXN],dist[MAXN],curedge[MAXN],pre[MAXN];
    int cur_flow,max_flow,u,tmp,neck,i;
    memset(dist,0,sizeof(dist));
    memset(numb,0,sizeof(numb));
    for(i = 1 ; i <= nv ; ++i)
        curedge[i] = net[i];
    numb[nv] = nv;
    max_flow = 0;
    u = s;
    while(dist[s] < nv){
        if(u == t){
            cur_flow = INF+1;
            for(i = s; i != t;i = edge[curedge[i]].v) 
                if(cur_flow > edge[curedge[i]].cap){
                    neck = i;
                    cur_flow = edge[curedge[i]].cap;
                }
            for(i = s; i != t; i = edge[curedge[i]].v)
            {
                tmp = curedge[i];
                edge[tmp].cap -= cur_flow;
                edge[tmp].flow += cur_flow;
                tmp = edge[tmp].pair;
                edge[tmp].cap += cur_flow;
                edge[tmp].flow -= cur_flow;
            }
            max_flow += cur_flow;
            u = neck;
        }
        for(i = curedge[u]; i != -1; i = edge[i].next)
            if(edge[i].cap > 0 && dist[u] == dist[edge[i].v]+1)
                break;
        if(i != -1){
            curedge[u] = i;
            pre[edge[i].v] = u;
            u = edge[i].v;
        }else{
            if(0 == --numb[dist[u]]) break;
            curedge[u] = net[u];
            for(tmp = nv,i = net[u]; i != -1; i = edge[i].next)
                if(edge[i].cap > 0)
                    tmp = tmp<dist[edge[i].v]?tmp:dist[edge[i].v];
            dist[u] = tmp + 1;
            ++numb[dist[u]];
            if(u != s) u = pre[u];
        }
    }
    return max_flow;
}

int main()
{
    int i,j,k,h,time,ans,n,work[7],day,week;
    scanf("%d",&time);
    while(time--)
    {
        init();
        scanf("%d",&n);
        for(i = 1; i <= n; ++i)
        {
            for(j = 0;j < 7; ++j)
                scanf("%d",&work[j]);
            scanf("%d%d",&day,&week);
            add_edge(s,i,day);
            for(j = 0;j < week; ++j)
                for(k = 0;k < 7; ++k)
                    if(work[k]) // can do 
                        add_edge(i,n+j*7+k+1,INF);
            max_week = max_week>week?max_week:week;
            sum += day;
        }
        t = n+max_week*7+1;
        nv = t + 1;
        for(i = 0;i < max_week; ++i )
            for(j = 1;j <= 7; ++j)
                add_edge(n+i*7+j,t,1);
        ans = ISAP();
        if(sum == ans)
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}