POJ1149 PIGS

PIGS
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 21217 Accepted: 9686
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 3
3 1 10
2 1 2 2
2 1 3 3
1 2 6
Sample Output

7
Source

Croatia OI 2002 Final Exam - First day

WA:

/*WA的厉害，浪费时间太多了，换了一个矩阵的不调了 路过大神求指点 */ #include<iostream>#include<cstring>#include<cstdio>#include<cstdlib>#include<queue>using namespace std;const int MAXN = 1200 ;#define MAXE MAXN*MAXN*2#define INF 105000000int head[MAXN];struct Edge{ int to,flow,next; }e[MAXE];int tot,S,T,dep[MAXN],n,m,pig[MAXN];//m--Pig Houseint now[MAXN];void Init(){    tot=0;    memset(head,-1,sizeof head );}queue<int> q;void Insert(int u,int v,int flow){    e[tot].to=v;e[tot].next=head[u];    e[tot].flow=flow;head[u]=tot++;    e[tot].to=u;e[tot].next=head[v];    e[tot].flow=0;head[v]=tot++;}bool BFS(int S,int T){    memset(dep,-1,sizeof dep );    dep[S]=0;q.push(S);    while(!q.empty()){        int u=q.front();q.pop();        for(int i=head[u];i!=-1;i=e[i].next){            int v=e[i].to,w=e[i].flow;            if(w&&dep[v]==-1){                dep[v]=dep[u]+1;                q.push(v);            }        }    }    return dep[T]!=-1;}int DFS(int u,int Flow){    int rt=0;    if(u==T) return Flow;    for(int i=head[u];i!=-1;i=e[i].next){        int v=e[i].to,w=e[i].flow;        if(dep[v]==dep[u]+1&&w){            int x=DFS(v,min(Flow-rt,w));            e[i].flow-=x;e[i^1].flow+=x;            rt+=x;Flow-=x;        }    }    return rt;}int Dinic(int S,int T){    int ans=0;    while(BFS(S,T)) ans+=DFS(S,INF);    return ans;}int main(){    while(~scanf("%d%d",&m,&n)){        T=n+1;        Init();        for(int i=1;i<=m;i++) scanf("%d",&pig[i]);        memset(now,0,sizeof now );        for(int x,i=1;i<=n;i++){            scanf("%d",&x);//鐚垗鐨勯挜鍖欐暟鐩?            for(int y,j=1;j<=x;j++){                scanf("%d",&y);                if(now[y]) Insert(now[y],i,INF);                else Insert(S,i,pig[y]);                now[y]=i;            }            scanf("%d",&x);//涔扮尓鏁伴噺            Insert(i,T,x);        }        int ans=Dinic(S,T);        printf("%d\n",ans);    }    return 0;}

AC :

#include <cstdio>#include <iostream>#include <cstring>#include <queue>using namespace std;const int INF = (1<<30);int pig[1100],flag[1100],res[110][110],dis[110];int m,n;    //m猪圈，n个人int S = 0,T ;void init(){    for(int i = 1 ; i <= m ;i++)        scanf("%d",&pig[i]);    memset(res,0,sizeof(res));    memset(flag,0,sizeof(flag));    for(int i = 1 ; i <= n ;i++){        int num;        scanf("%d",&num);        for(int j = 0 ; j < num ;j++){            int room;            scanf("%d",&room);            if(flag[room] == 0) res[S][i] += pig[room];            else res[flag[room]][i] = INF;            flag[room] = i;        }       int xu;       scanf("%d",&xu);       res[i][T] = xu;    }}int bfs(){    int tmp;    queue<int>q;    memset(dis,-1,sizeof(dis));    dis[T] = 0;    q.push(T);    while(!q.empty()){       tmp = q.front();       if(tmp == S) return 1;       q.pop();       for(int i = 0 ; i <= T;i++){        if(dis[i] == -1&&res[i][tmp]){            dis[i] = dis[tmp]+1;            q.push(i);        }       }    }    return 0;}int dfs(int cur,int c){    if(cur == T)    return c;    int tmp = c,flow;    for(int i = 0 ; i <= T ;i++){        if(tmp == 0)    break;        if(dis[i]+1 == dis[cur]&&res[cur][i]){            flow = dfs(i,min(res[cur][i],tmp));            res[cur][i] -= flow;            res[i][cur] += flow;            tmp -= flow;        }    }    return c-tmp;}void dinic(){    int maxflow = 0;    while(bfs()) maxflow+=dfs(S,INF);    printf("%d\n",maxflow);}int main(){    while(~scanf("%d%d",&m,&n)){       T = n+1;       init();       dinic();    }    return 0;}