PIGS
| Time Limit: 1000MS | Memory Limit: 10000K | |
Description
Mirko works on a pig farm that consists of M locked pig-houses and Mirko can't unlock any pighouse because he doesn't have the keys. Customers come to the farm one after another. Each of them has keys to some pig-houses and wants to buy a certain number of pigs. All data concerning customers planning to visit the farm on that particular day are available to Mirko early in the morning so that he can make a sales-plan in order to maximize the number of pigs sold. More precisely, the procedure is as following: the customer arrives, opens all pig-houses to which he has the key, Mirko sells a certain number of pigs from all the unlocked pig-houses to him, and, if Mirko wants, he can redistribute the remaining pigs across the unlocked pig-houses. An unlimited number of pigs can be placed in every pig-house. Write a program that will find the maximum number of pigs that he can sell on that day.
Input
The first line of input contains two integers M and N, 1 <= M <= 1000, 1 <= N <= 100, number of pighouses and number of customers. Pig houses are numbered from 1 to M and customers are numbered from 1 to N. The next line contains M integeres, for each pig-house initial number of pigs. The number of pigs in each pig-house is greater or equal to 0 and less or equal to 1000. The next N lines contains records about the customers in the following form ( record about the i-th customer is written in the (i+2)-th line): A K1 K2 ... KA B It means that this customer has key to the pig-houses marked with the numbers K1, K2, ..., KA (sorted nondecreasingly ) and that he wants to buy B pigs. Numbers A and B can be equal to 0.
Output
The first and only line of the output should contain the number of sold pigs.
Sample Input
3 33 1 102 1 2 22 1 3 31 2 6
Sample Output
7
Source
题目大意：
Mirko 养着一些猪, 猪关在一些猪圈里面, 猪圈是锁着的他自己没有钥匙 (汗) 只有要来买猪的顾客才有钥匙, 顾客依次来每个顾客会用他的钥匙打开一些猪圈买 走一些猪, 然后锁上. 在锁上之前 Mirko 有机会重新分配这几个已打开猪圈的猪, 现在给出一开始每个猪 圈的猪数, 每个顾客所有的钥匙和要买走的猪数问 Mirko 最多能卖掉几头猪
以样例输入为例
有m个猪圈，n个顾客
接下来一行有m个数，表示每个猪圈开始猪的数量
接下来n行，
每行第一个数表示第i个顾客手里的钥匙数p，后面p个数表示是哪个猪圈的钥匙，最后一个数表示他要买的猪的数量
最大流
很容易想到的建图方法
上图修改如下：
但是这样会有100000左右个点，TLE
所以我们要对点进行合并
合并规则：
规律 1. 如果几个节点的流量的来源完全相同，则可以把它们合并成一个。
规律 2. 如果几个节点的流量的去向完全相同，则可以把它们合并成一个。 规律 3. 如果从点 u 到点 v 有一条容量为 +∞ 的边，并且 u 是 v 的唯一流量来源，或者 v 是 u 的唯一流量去向，则可以把 u 和 v 合并成一个节点。
所以最终构图方式：
1、由源点向每个猪圈第一个要访问的顾客连一条边，流量为一开始猪的数量，如果源点向同一个顾客有多条边，就把他们合并为1条边，流量相加。
2、如果顾客j在顾客i后访问猪圈k，则由i向j连一条流量为inf的边
3、每个顾客向汇点连一条边，流量为他要买的猪的数量
更详细的解释推荐博客
#include#include #include #include #include #define inf 600001using namespace std;int n,m,qq,tot=1,ans;int src,decc;int front[112],cap[22001],lev[112],cnt[112],nextt[22001],to[22001];queue q;int pre[1002],buy[112],kai[112][112],last[1002];void add(int u,int v,int w){ to[++tot]=v;cap[tot]=w;nextt[tot]=front[u];front[u]=tot; to[++tot]=u;cap[tot]=0;nextt[tot]=front[v];front[v]=tot;}bool bfs(){ for(int i=0;i<=m+1;i++) {cnt[i]=front[i];lev[i]=-1;} while(!q.empty()) q.pop(); q.push(src);lev[src]=0; while(!q.empty()) { int now=q.front();q.pop(); for(int i=front[now];i!=0;i=nextt[i]) { int t=to[i]; if(cap[i]>0&&lev[t]==-1) { q.push(t); lev[t]=lev[now]+1; if(t==decc) return true; } } } return false;}int dinic(int now,int flow){ if(now==decc) return flow; int delta,rest=0; for(int & i=cnt[now];i!=0;i=nextt[i]) { int t=to[i]; if(lev[t]==lev[now]+1&&cap[i]>0) { delta=dinic(t,min(cap[i],flow-rest)); if(delta) { cap[i]-=delta;cap[i^1]+=delta; rest+=delta;if(rest==flow) break; } } } if(rest!=flow) lev[now]=-1; return rest;}int main(){ scanf("%d%d",&n,&m); int p,x; for(int i=1;i<=n;i++) scanf("%d",&pre[i]); for(int i=1;i<=m;i++) { scanf("%d",&p); for(int j=1;j<=p;j++) { scanf("%d",&x); if(!last[x]) { kai[0][i]+=pre[x]; last[x]=i; } else kai[last[x]][i]=inf; } scanf("%d",&buy[i]); } decc=m+1; for(int i=0;i<=m;i++) for(int j=0;j<=m;j++) if(kai[i][j]) add(i,j,kai[i][j]); for(int i=1;i<=m;i++) add(i,decc,buy[i]); while(bfs()) ans+=dinic(src,inf); printf("%d",ans);}