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
*/