UVA 1345 - Jamie's Contact Groups（二分+最大流）

UVA 1345 - Jamie's Contact Groups

题目链接

题意：给定一些人，每个人有一个分组，现在要每个人选一个分组，使得所有分组中最大的人数最小，问这个最小值是多少

思路：二分答案，然后利用最大流去判定，源点往每个人建一条边容量为1，每个人往各自的分组建一条边，容量为1，分组向汇点建一条边，容量为二分出来的值，这样跑一下最大流如果最大流等于n，就是能满足

代码：

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#include <map>
#include <string>
#include <iostream>
#include <set>
using namespace std;

const int MAXNODE = 1505;
const int MAXEDGE = 1100005;

typedef int Type;
const Type INF = 0x3f3f3f3f;

struct Edge {
int u, v;
Type cap, flow;
Edge() {}
Edge(int u, int v, Type cap, Type flow) {
this->u = u;
this->v = v;
this->cap = cap;
this->flow = flow;
}
};

struct Dinic {
int n, m, s, t;
Edge edges[MAXEDGE];
int first[MAXNODE];
int next[MAXEDGE];
bool vis[MAXNODE];
Type d[MAXNODE];
int cur[MAXNODE];
vector<int> cut;

void init(int n) {
this->n = n;
memset(first, -1, sizeof(first));
m = 0;
}
void add_Edge(int u, int v, Type cap) {
edges[m] = Edge(u, v, cap, 0);
next[m] = first[u];
first[u] = m++;
edges[m] = Edge(v, u, 0, 0);
next[m] = first[v];
first[v] = m++;
}

bool bfs() {
memset(vis, false, sizeof(vis));
queue<int> Q;
Q.push(s);
d[s] = 0;
vis[s] = true;
while (!Q.empty()) {
int u = Q.front(); Q.pop();
for (int i = first[u]; i != -1; i = next[i]) {
Edge& e = edges[i];
if (!vis[e.v] && e.cap > e.flow) {
vis[e.v] = true;
d[e.v] = d[u] + 1;
Q.push(e.v);
}
}
}
return vis[t];
}

Type dfs(int u, Type a) {
if (u == t || a == 0) return a;
Type flow = 0, f;
for (int &i = cur[u]; i != -1; i = next[i]) {
Edge& e = edges[i];
if (d[u] + 1 == d[e.v] && (f = dfs(e.v, min(a, e.cap - e.flow))) > 0) {
e.flow += f;
edges[i^1].flow -= f;
flow += f;
a -= f;
if (a == 0) break;
}
}
return flow;
}

bool Maxflow(int s, int t, int tot) {
this->s = s; this->t = t;
Type flow = 0;
while (bfs()) {
for (int i = 0; i < n; i++)
cur[i] = first[i];
flow += dfs(s, INF);
}
return flow == tot;
}

void MinCut() {
cut.clear();
for (int i = 0; i < m; i += 2) {
if (vis[edges[i].u] && !vis[edges[i].v])
cut.push_back(i);
}
}
} gao;

const int N = 1005;
const int M = 505;

int n, m, cnt[M];
vector<int> f
;
char str[100005];

bool judge(int c) {
gao.init(n + m + 2);
int s = 0, t = n + m + 1;
for (int i = 0; i < n; i++) {
gao.add_Edge(0, i + 1, 1);
for (int j = 0; j < f[i].size(); j++) {
gao.add_Edge(i + 1, f[i][j] + n + 1, 1);
}
}
for (int i = 0; i < m; i++)
gao.add_Edge(n + i + 1, t, c);
return gao.Maxflow(s, t, n);
}

int main() {
while (~scanf("%d%d", &n, &m) && n || m) {
memset(cnt, 0, sizeof(cnt));
while ((getchar()) != '\n');
for (int i = 0; i < n; i++) {
f[i].clear();
gets(str);
int len = strlen(str);
int sum = 0;
int flag = 0;
for (int j = 0; j < len; j++) {
if (str[j] >= '0' && str[j] <= '9') {
flag = 1;
sum = sum * 10 + str[j] - '0';
} else {
if (flag) {
f[i].push_back(sum);
cnt[sum]++;
}
flag = 0;
sum = 0;
}
}
if (flag) {
f[i].push_back(sum);
cnt[sum]++;
}
}
int l = 0, r = 0;
for (int i = 0; i < m; i++) r = max(r, cnt[i]);
while (l < r) {
int mid = (l + r) / 2;
if (judge(mid)) r = mid;
else l = mid + 1;
}
printf("%d\n", l);
}
return 0;
}