小豬分配 , 最大流
阿新 • • 發佈:2018-11-09
add rom some som max rds initial else RoCE
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
Sample Output
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
7
題意 : 有N個顧客,有M個豬圈,每個豬圈有一定的豬,在開始的時候豬圈都是關閉的,
顧客來買豬,顧客打開某個豬圈,可以在其中挑選一定的豬的數量,在這個顧客走後,可以在打開的
豬圈中將某個豬圈的一些豬牽到另外一個打開的豬圈,然後所有的豬圈會關閉,這樣下一個顧客來了繼續上面的工作
思路分析 :
參考此文章即可 :http://wenku.baidu.com/view/0ad00abec77da26925c5b01c.html
代碼示例 :
const int maxn = 1e4+5; const int inf = 0x3f3f3f3f; int m, n; int a[1005]; vector<int>ve[1005]; struct node { int to, flow; int next; }e[maxn]; int head[maxn]; int cnt = 0; int s, t; void addedge(int u, int v, int w){ e[cnt].next = head[u], e[cnt].to = v, e[cnt].flow = w, head[u] = cnt++; e[cnt].next = head[v], e[cnt].to = u, e[cnt].flow = 0, head[v] = cnt++; } int belong[1005]; int dep[1005], que[maxn]; bool bfs(int s, int t){ int head1 = 0, tail = 1; memset(dep, 0, sizeof(dep)); que[0] = s; dep[s] = 1; while(head1 < tail){ int v = que[head1++]; for(int i = head[v]; i != -1; i = e[i].next){ int to = e[i].to; if (e[i].flow && !dep[to]) { dep[to] = dep[v]+1; que[tail++] = to; } } } return dep[t]; } int dfs(int u, int f1){ if (f1 == 0 || u == t) return f1; int f = 0; for(int i = head[u]; i != -1; i = e[i].next){ int to = e[i].to; if (e[i].flow && dep[to] == dep[u]+1){ int x = dfs(to, min(e[i].flow, f1)); e[i].flow -= x; e[i^1].flow += x; f1 -= x, f += x; if (f1 == 0) return f; } } if (!f) dep[u] = -2; return f; } int maxflow(int s, int t){ int ans = 0; while(bfs(s, t)){ ans += dfs(s, inf); } return ans; } int main() { //freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); int x, y; cin >> m >> n; s = 0, t = n+1; memset(head, -1, sizeof(head)); for(int i = 1; i <= m; i++) scanf("%d", &a[i]); for(int i = 1; i <= n; i++){ scanf("%d", &x); for(int j = 1; j <= x; j++) { scanf("%d", &y); ve[i].push_back(y); } scanf("%d", &y); addedge(i, t, y); } for(int i = 1; i <= n; i++){ for(int j = 0; j < ve[i].size(); j++){ int v = ve[i][j]; if (!belong[v]) { belong[v] = i; addedge(s, i, a[v]); } else { addedge(belong[v], i, inf); belong[v] = i; } } } printf("%d\n", maxflow(s, t)); return 0; }
小豬分配 , 最大流