【网络流】：poj1698，Alice's Chance

题目大意：爱丽丝要拍电影，有n部电影，规定爱丽丝每部电影在每个礼拜只有固定的几天可以拍电影，只可以拍前面w个礼拜，并且这部电影要拍d天,问爱丽丝能不能拍完所有的电影
第一行代表有多少组数据
对于每组数据第一行代表有n部电影
接下来2到n+1行，每行代表一个电影，每行9个数，前面7个数，1代表拍，0代表不拍，第8个数代表要拍几天，第9个数代表有几个礼拜时间拍

源点：0号点为源点；

week点：1到350（7*50）号点 分别表示50个星期的每一天；

film点：351到370号点 表示20个film；

汇点：371号点表示汇点。

源点到每个week点连一个边，容量为1；

每个film点到汇点连一个边，容量为d，表示该film需要拍摄的天数；

week点与film点之间如果有对应关系，则连一条边，容量为1。连边时注意代码的顺序：

while(--w>=0)
            {
                for(k=1;k<=7;k++)
                {
                    if(a[k]==1)
                    {
                        add_edge(w*7+k,350+j,1,0);
                    }
                }
            }
            /*
            for(k=1;k<=7;k++)
            {
                if(a[k]==1)
                {
                    while(--w>=0)
                    {
                        add_edge(w*7+k,350+j,1,0);
                    }
                }
            }
            */

/*
 * Dinic algo for max flow
 *
 * This implementation assumes that #nodes, #edges, and capacity on each edge <= INT_MAX,
 * which means INT_MAX is the best approximation of INF on edge capacity.
 * The total amount of max flow computed can be up to LLONG_MAX (not defined in this file),
 * but each 'dfs' call in 'dinic' can return <= INT_MAX flow value.
 */
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#include <string.h>
#include <assert.h>
#include <queue>
#include <vector>

# include<iostream>
# include<cstring>
# include<map>

#define N (380)       //==================make sure this is the total node number!
#define M (N*N+4*N)

typedef long long LL;

using namespace std;

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

int head[N], level[N], cur[N];
int num_of_edges;

//When there are multiple test sets, you need to re-initialize before each
void dinic_init(void)
{
    num_of_edges = 0;
    memset(head, -1, sizeof(head));
    return;
}

int add_edge(int u, int v, int c1, int c2)
{
    int& i=num_of_edges;

    assert(c1>=0 && c2>=0 && c1+c2>=0); // check for possibility of overflow
    e[i].v = v;
    e[i].cap = c1;
    e[i].next = head[u];
    head[u] = i++;

    e[i].v = u;
    e[i].cap = c2;
    e[i].next = head[v];
    head[v] = i++;
    return i;
}

void print_graph(int n)
{
    for (int u=0; u<n; u++)
    {
        printf("%d: ", u);
        for (int i=head[u]; i>=0; i=e[i].next)
        {
            printf("%d(%d)", e[i].v, e[i].cap);
        }
        printf("\n");
    }
    return;
}

//Find all augmentation paths in the current level graph This is the recursive version
int dfs(int u, int t, int bn)
{
    if (u == t) return bn;
    int left = bn;
    for (int i=head[u]; i>=0; i=e[i].next)
    {
        int v = e[i].v;
        int c = e[i].cap;
        if (c > 0 && level[u]+1 == level[v])
        {
            int flow = dfs(v, t, min(left, c));
            if (flow > 0)
            {
                e[i].cap -= flow;
                e[i^1].cap += flow;
                cur[u] = v;
                left -= flow;
                if (!left) break;
            }
        }
    }
    if (left > 0) level[u] = 0;
    return bn - left;
}

bool bfs(int s, int t)
{
    memset(level, 0, sizeof(level));
    level[s] = 1;
    queue<int> q;
    q.push(s);
    while (!q.empty())
    {
        int u = q.front();
        q.pop();
        if (u == t) return true;
        for (int i=head[u]; i>=0; i=e[i].next)
        {
            int v = e[i].v;
            if (!level[v] && e[i].cap > 0)
            {
                level[v] = level[u]+1;
                q.push(v);
            }
        }
    }
    return false;
}

LL dinic(int s, int t)
{
    LL max_flow = 0;

    while (bfs(s, t))
    {
        memcpy(cur, head, sizeof(head));
        max_flow += dfs(s, t, INT_MAX);
    }
    return max_flow;
}

int upstream(int s, int n)
{
    int cnt = 0;
    vector<bool> visited(n);
    queue<int> q;
    visited[s] = true;
    q.push(s);
    while (!q.empty())
    {
        int u = q.front();
        q.pop();
        for (int i=head[u]; i>=0; i=e[i].next)
        {
            int v = e[i].v;
            if (e[i].cap > 0 && !visited[v])
            {
                visited[v] = true;
                q.push(v);
                cnt++;
            }
        }
    }
    return cnt; // excluding s
}


int main()
{
    int t,n,i,j,k,a[8],d,w,sum;

    cin>>t;

    for(i=1;i<=t;i++)
    {
        dinic_init();

        sum=0;
        for(j=1;j<=350;j++)
        {
            add_edge(0,j,1,0);
        }

        cin>>n;
        for(j=1;j<=n;j++)
        {
            memset(a,0,sizeof(a));
            for(k=1;k<=7;k++)
            {
                cin>>a[k];
            }
            cin>>d>>w;
            sum+=d;
            while(--w>=0)
            {
                for(k=1;k<=7;k++)
                {
                    if(a[k]==1)
                    {
                        add_edge(w*7+k,350+j,1,0);
                    }
                }
            }
            /*
            for(k=1;k<=7;k++)
            {
                if(a[k]==1)
                {
                    while(--w>=0)
                    {
                        add_edge(w*7+k,350+j,1,0);
                    }
                }
            }
            */
            add_edge(350+j,371,d,0);
        }

        if(sum==dinic(0,371))
        {
            cout<<"Yes"<<endl;
        }
        else
        {
            cout<<"No"<<endl;
        }
    }


    return 0;
}