网络流二十四题之星际转移问题

这道题的建模是按照天数建模的分层模型，然后枚举需要多少天，需要对标号进行合理的设计。
1.每一天的地球，月球分别向源点，汇点连一条边，容量inf
2.对于n个空间站，向他的下一天连一条边，容量无限
.3.如果有一天飞船在前一天在i,这一天在j那么将这两个点连接在一起。
然后每天判断是否满流

#include <cstdio>
#include <iostream>
#include <cstring>
#include <vector>
#include <algorithm>
#include <queue>
#include <map>
using namespace std;
const int MAXN = 100010;//点数的最大值
const int MAXM = 400010;//边数的最大值
const int INF = 0x3f3f3f3f;
struct Edge
{
    int from,to,next,cap,flow;
} edge[MAXM]; //注意是MAXM
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],cur[MAXN];
void init()
{
    tol = 0;
    memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w,int rw = 0)
{
    edge[tol].from = u;
    edge[tol].to = v;
    edge[tol].cap = w;
    edge[tol].flow = 0;
    edge[tol].next = head[u];
    head[u] = tol++;
    edge[tol].from = v;
    edge[tol].to = u;
    edge[tol].cap = rw;
    edge[tol].flow = 0;
    edge[tol].next = head[v];
    head[v] = tol++;
}
int Q[MAXN];
void BFS(int start,int end)
{
    memset(dep,-1,sizeof(dep));
    memset(gap,0,sizeof(gap));
    gap[0] = 1;
    int front = 0, rear = 0;
    dep[end] = 0;
    Q[rear++] = end;
    while(front != rear)
    {
        int u = Q[front++];
        for(int i = head[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].to;
            if(dep[v] != -1)
                continue;
            Q[rear++] = v;
            dep[v] = dep[u] + 1;
            gap[dep[v]]++;
        }
    }
}
int S[MAXN];
int sap(int start,int end,int N)
{
    BFS(start,end);
    memcpy(cur,head,sizeof(head));
    int top = 0;
    int u = start;
    int ans = 0;
    while(dep[start] < N)
    {
        if(u == end)
        {
            int Min = INF;
            int inser;
            for(int i = 0; i < top; i++)
                if(Min > edge[S[i]].cap - edge[S[i]].flow)
                {
                    Min = edge[S[i]].cap - edge[S[i]].flow;
                    inser = i;
                }
            for(int i = 0; i < top; i++)
            {
                edge[S[i]].flow += Min;
                edge[S[i]^1].flow -= Min;
            }
            ans += Min;
            top = inser;
            u = edge[S[top]^1].to;
            continue;
        }
        bool flag = 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])
            {
                flag = true;
                cur[u] = i;
                break;
            }
        }
        if(flag)
        {
            S[top++] = cur[u];
            u = v;
            continue;
        }
        int Min = N;
        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 != start)
            u = edge[S[--top]^1].to;
    }
    return ans;
}
void print_all()
{
    for(int i = 0; i<tol; i+=2)
    {
        printf("%d->%d cap = %d flow = %d\n",edge[i].from,edge[i].to,edge[i].cap,edge[i].from);
    }
}
int a[1001][1001];
int h[1001];
int len[1001];
int main()
{
    int n,m,k;
    scanf("%d%d%d",&n,&m,&k);
    init();
    n += 2;
    for(int i = 1; i<=m; i++)
    {
        scanf("%d%d",&h[i],&len[i]);
        for(int j = 0; j<len[i]; j++)
        {
            scanf("%d",&a[i][j]);
            a[i][j] += 2;
        }
    }
    int s = 0,t = 9999;
    for(int time = 0; time <= 500; time ++)
    {
        addedge(s, time*n+2, INF);
        addedge(time*n+1,t,INF);
        if(time != 0)
        {
            for(int i  =1; i<=m; i++)
            {
                int x = a[i][(time -1) % len[i]];
                int y = a[i][time % len[i]];
                addedge((time - 1) * n + x,time *n + y,h[i]);
            }
            for(int i = 1; i<=n; i++)
            {
                addedge((time-1)*n+ i,time*n+i,INF);
            }

        }
        int ans = sap(s,t,2000);
        //print_all();
        //printf("%d\n\n",ans);
        //puts("");
        if(ans >= k)
        {
            printf("%d\n",time);
            break;
        }
    }
    return 0;
}
/*
2 2 1
1 3 0 1 2
1 3 1 2 -1
*/