【二分图多重匹配 && Dinic】POJ - 2289 Jamie's Contact Groups

Problem Description

给你m个组别，每个组别可以只能放特定的人。并最大限度地减少最大组的人数

思路:

0-(m-1)代表人，m代表超级源点，n-(n+m)代表组别，n+m+1代表超级汇点。二分 人到超级汇点的流量，最大流求最大匹配，如果最大匹配 = n就减少流量。

#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
#define N 1005
#define M 505
#define inf 0x3f3f3f3f
struct node
{
int to, cap, next;
};
node Map[2*N*M];
vector<int> a
;
int n, m, cnt;
int head[N+M], vis[N+M];
bool bfs(int s, int e)
{
memset(vis, -1, sizeof(vis));
queue<int> q;
q.push(s);
vis[s] = 0;
while(!q.empty())
{
s = q.front(), q.pop();
for(int i = head[s]; ~i; i = Map[i].next)
{
int to = Map[i].to, cap = Map[i].cap;
if(cap && vis[to] == -1)
{
vis[to] = vis[s] + 1;
q.push(to);
}
}
}
if(vis[e] == -1) return 0;
else return 1;
}
int dfs(int s, int e, int f)
{
if(s == e) return f;
int ans = 0;
for(int i = head[s]; ~i; i = Map[i].next)
{
int to = Map[i].to, &cap = Map[i].cap;
if(vis[to] > vis[s] && cap)
{
int d = dfs(to, e, min(f, cap));
if(d)
{
cap -= d;
Map[i^1].cap += d;
f -= d;
ans += d;
if(!f) break;
}
}
}
if(ans) return ans;
vis[s] = -1;
return 0;
}
int dinic(int s, int e)
{
int ans = 0;
while(bfs(s, e))
{
ans += dfs(s, e, inf);
}
return ans;
}
void add(int u, int v, int cap)
{
Map[cnt].to = v;
Map[cnt].cap = cap;
Map[cnt].next = head[u];
head[u] = cnt++;
Map[cnt].to = u;
Map[cnt].cap = 0;
Map[cnt].next = head[v];
head[v] = cnt++;
}
void build(int mid)//m是源点 n+m+1是汇点
{
int i, j;
memset(head, -1, sizeof(head));
cnt = 0;
for(i = 1; i <= n; i++)
{
add(m, i+m, 1);//超级源点->组别
for(j = 0; j < a[i].size(); j++)
{
add(i+m, a[i][j], 1);//组别->人
}
}
for(i = 0; i < m; i++)
{
add(i, n+m+1, mid);//人->超级汇点
}
}
int main()
{
char s[20];
int num;
while(~scanf("%d %d", &n, &m) && n+m)
{
for(int i = 1; i <= n; i++)
{
a[i].clear();
scanf("%s", s);
while(getchar() != '\n')
{
scanf("%d", &num);
a[i].push_back(num);
}
}
int l = 1, r = n;
while(l <= r)
{
int mid = (l + r) / 2;
build(mid);//对于每个人到超级汇点的流量，重新建边
if(dinic(m, n+m+1) == n) r = mid - 1;//求最大流
else l = mid + 1;
}
printf("%d\n", r + 1);
}
return 0;
}