poj1274 The Perfect Stall(二分图匹配 / 最大流)
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The Perfect Stall
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 20547 Accepted: 9263
Description
Farmer John completed his new barn just last week, complete with all the latest milking technology. Unfortunately, due to engineering problems, all the stalls in the new barn are different. For the first week, Farmer John randomly assigned cows to stalls, but it quickly became clear that any given cow was only willing to produce milk in certain stalls. For the last week, Farmer John has been collecting data on which cows are willing to produce milk in which stalls. A stall may be only assigned to one cow, and, of course, a cow may be only assigned to one stall.
Given the preferences of the cows, compute the maximum number of milk-producing assignments of cows to stalls that is possible.
Given the preferences of the cows, compute the maximum number of milk-producing assignments of cows to stalls that is possible.
Input
The input includes several cases. For each case, the first line contains two integers, N (0 <= N <= 200) and M (0 <= M <= 200). N is the number of cows that Farmer John has and M is the number of stalls in the new barn. Each of the following N lines corresponds to a single cow. The first integer (Si) on the line is the number of stalls that the cow is willing to produce milk in (0 <= Si <= M). The subsequent Si integers on that line are the stalls in which that cow is willing to produce milk. The stall numbers will be integers in the range (1..M), and no stall will be listed twice for a given cow.
Output
For each case, output a single line with a single integer, the maximum number of milk-producing stall assignments that can be made.
Sample Input
5 52 2 53 2 3 42 1 53 1 2 51 2
Sample Output
4
题意:n头牛,m个位置,每头牛可以待在s个位置上,给每个牛分配一个位置,求问最多几头牛可以分配到位置。
这是二分图匹配的基础题,也可以用最大流来做,直接贴代码。。
二分图匹配:
#include <cstdio>#include <iostream>#include <algorithm>#include <cstring>#define maxn 205using namespace std;int maps[maxn][maxn], vis[maxn], link[maxn];int n, m;bool Find(int x){ for(int i = 1; i <= m; i++) { if(vis[i] == 0 && maps[x][i] == 1) { vis[i] = 1; if(link[i] == 0 || Find(link[i])) { link[i] = x; return true; } } } return false;}int main(){ int x, y; while(scanf("%d%d", &n, &m) != EOF) { memset(maps, 0, sizeof(maps)); memset(link, 0, sizeof(link)); for(int i = 1; i <= n; i++) { scanf("%d", &x); for(int j = 0; j < x; j++) { scanf("%d", &y); maps[i][y] = 1; } } int ans = 0; for(int i = 1; i <= n; i++) { memset(vis, 0, sizeof(vis)); if(Find(i)) ans++; } printf("%d\n", ans); }}
网络流最大流:
#include <cstdio>#include <iostream>#include <algorithm>#include <cstring>#define maxn 505using namespace std;int n, m;int maps[maxn][maxn], q[maxn * maxn];int d[maxn];//记录每个点所在的层int BFS()//BFS分层,判断是否存在增广路{ memset(d, -1, sizeof(d)); d[0] = 0; int f, r; f = 1; r = 1; q[r++] = 0; while(f < r) { int x = q[f++]; for(int i = 0; i <= n + m + 1; i++) { if(d[i] < 0 && maps[x][i] > 0) { d[i] = d[x] + 1; q[r++] = i; } } } if(d[n + m + 1] > 0) return 1; else return 0;}int Find(int x, int v)//找出某一条增广路的最大流{ int a; if(x == n + m + 1) return v; for(int i = 0; i <= n + m + 1; i++) { if(maps[x][i] > 0 && d[i] == d[x] + 1 && (a = Find(i, min(v, maps[x][i]))))//x到i有流量 且 i是x的下一层 且 i到汇点存在最大流 { maps[x][i] -= a;//增广路 maps[i][x] += a;//回退边 return a; } } return 0;}int main(){ int x, y, t; while(scanf("%d%d", &n, &m) != EOF) { memset(maps, 0, sizeof(maps)); for(int i = 1; i <= n; i ++) { scanf("%d", &x); for(int j = 1; j <= x; j ++) { scanf("%d", &y); maps[i][n + y] = 1; } } for(int i = 1; i <= n; i++) maps[0][i] = 1; for(int i = n + 1; i <= n + m; i++) maps[i][n + m + 1] = 1;// for(int i = 0; i <= n + m + 1; i++)// {// for(int j = 0; j <= n + m + 1; j++)// {// printf("%d ",maps[i][j]);// }// printf("\n");// } int ans = 0; while(BFS())//BFS搜索判断是否有从源点到汇点的通路 { while(t = Find(0, 2000000005))//查找增广路,求其最大流 ans += t; } printf("%d\n", ans); }}
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