UVa 1345 - Jamie's Contact Groups

链接

传送门

题意

有$n, (1 \leq n \leq 1000)$个人,$m, (1 \leq m \leq 500)$个组,一个人可能属于多个组。现在从一些组中删除一部分人,使得每个人仅在一个组中,且包含人数最多的组的人数最少。

思路

源点向每个人连一条容量为1的边,人和组之间连容量为1的边,组与汇点之间的边的容量为最大人数。二分最大人数,判断最大流是否等于$n$。

代码

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#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
typedef long long LL;
const int inf = 0x3f3f3f3f;
const int maxn = 1510;
struct Edge {
int from, to, cap, flow;
Edge() {}
Edge(int u, int v, int c, int f): from(u), to(v), cap(c), flow(f) {}
};
struct ISAP {
int n, m, s, t;
vector<Edge> edges;
vector<int> G[maxn];
bool vis[maxn];
int p[maxn];
int d[maxn];
int num[maxn];
int cur[maxn];
void AddEdge(int from, int to, int cap) {
edges.push_back(Edge(from, to, cap, 0));
edges.push_back(Edge(to, from, 0, 0));
m = int(edges.size());
G[from].push_back(m - 2);
G[to].push_back(m - 1);
}
void ClearAll(int n) {
this -> n = n;
for (int i = 0; i < n; ++i) {
G[i].clear();
}
edges.clear();
}
bool BFS() {
memset(vis, 0, sizeof vis);
vis[t] = true;
d[t] = 0;
queue<int> Q;
Q.push(t);
while (!Q.empty()) {
int x = Q.front();
Q.pop();
for (int i = 0; i < G[x].size(); ++i) {
Edge& e = edges[G[x][i] ^ 1];
if (!vis[e.from] && e.cap > e.flow) {
vis[e.from] = true;
d[e.from] = d[x] + 1;
Q.push(e.from);
}
}
}
return vis[s];
}
int Augment() {
int x = t, a = inf;
while (x != s) {
Edge& e = edges[p[x]];
a = min(a, e.cap - e.flow);
x = e.from;
}
x = t;
while (x != s) {
edges[p[x]].flow += a;
edges[p[x] ^ 1].flow -= a;
x = edges[p[x]].from;
}
return a;
}
int Maxflow(int s, int t) {
this -> s = s, this -> t = t;
int flow = 0;
BFS();
memset(num, 0, sizeof num);
for (int i = 0; i < n; ++i) {
++num[d[i]];
}
memset(cur, 0, sizeof cur);
int x = s;
while (d[s] < n) {
if (x == t) {
flow += Augment();
x = s;
}
bool ok = false;
for (int i = cur[x]; i < G[x].size(); ++i) {
Edge& e = edges[G[x][i]];
if (e.cap > e.flow && d[x] == d[e.to] + 1) {
ok = true;
p[e.to] = G[x][i];
cur[x] = i;
x = e.to;
break;
}
}
if (!ok) {
int m = n - 1;
for (int i = 0; i < G[x].size(); ++i) {
Edge& e = edges[G[x][i]];
if (e.cap > e.flow) {
m = min(m, d[e.to]);
}
}
if (--num[d[x]] == 0) {
break;
}
++num[d[x] = m + 1];
cur[x] = 0;
if (x != s) {
x = edges[p[x]].from;
}
}
}
return flow;
}
void ClearFlow() {
for (int i = 0; i < m; ++i) {
edges[i].flow = 0;
}
}
} isap;
int main() {
int n, m;
while (~scanf("%d%d", &n, &m) && (n || m)) {
isap.ClearAll(n + m + 5);
for (int i = 0; i < n; ++i) {
scanf("%*s");
isap.AddEdge(n + m, i, 1);
while (getchar() != '\n') {
int x;
scanf("%d", &x);
isap.AddEdge(i, n + x, 1);
}
}
int L = 1, R = n, x = isap.m;
for (int i = 0; i < m; ++i) {
isap.AddEdge(n + i, n + m + 1, inf);
}
int ans = n;
while (L <= R) {
int M = L + (R - L) / 2;
isap.ClearFlow();
for (int i = x; i < isap.m; i += 2) {
isap.edges[i].cap = M;
}
if (isap.Maxflow(n + m, n + m + 1) == n) {
R = M - 1;
ans = min(ans, M);
} else {
L = M + 1;
}
}
printf("%d\n", ans);
}
return 0;
}

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