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Description
Presumably, you all have known the question of stable marriage match. A girl will choose a boy; it is similar as the game of playing house we used to play when we are kids. What a happy time as so many friends playing together. And it is normal that a fight or a quarrel breaks out, but we will still play together after that, because we are kids.
Now, there are 2n kids, n boys numbered from 1 to n, and n girls numbered from 1 to n. you know, ladies first. So, every girl can choose a boy first, with whom she has not quarreled, to make up a family. Besides, the girl X can also choose boy Z to be her boyfriend when her friend, girl Y has not quarreled with him. Furthermore, the friendship is mutual, which means a and c are friends provided that a and b are friends and b and c are friend.
Once every girl finds their boyfriends they will start a new round of this game―marriage match. At the end of each round, every girl will start to find a new boyfriend, who she has not chosen before. So the game goes on and on.
Now, here is the question for you, how many rounds can these 2n kids totally play this game?
Input
There are several test cases. First is a integer T, means the number of test cases.
Each test case starts with three integer n, m and f in a line (3<=n<=100,0
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#pragma comment(linker, "/STACK:102400000,102400000") //手动扩栈防RE
using namespace std;
const int PO = 205;
const int N = 500;
const int M = 50000;
const int INF = 0x3f3f3f3f;
typedef long long ll;
int n, m, k, s, t;
int f[PO];
struct Node{
int x, y;
}mg[M];
int vis[PO][PO];
int ec, head[N], first[N], que[N], lev[N];
int Next[M], to[M], v[M];
void init() {
ec = 0;
memset(first, -1, sizeof(first));
}
void addEdge(int a,int b,int c) {
to[ec] = b;
v[ec] = c;
Next[ec] = first[a];
first[a] = ec++;
to[ec] = a;
v[ec] = 0;
Next[ec] = first[b];
first[b] = ec++;
}
int BFS() {
int kid, now, f = 0, r = 1, i;
memset(lev, 0, sizeof(lev));
que[0] = s, lev[s] = 1;
while (f < r) {
now = que[f++];
for (i = first[now]; i != -1; i = Next[i]) {
kid = to[i];
if (!lev[kid] && v[i]) {
lev[kid] = lev[now] + 1;
if (kid == t) return 1;
que[r++] = kid;
}
}
}
return 0;
}
int DFS(int now, int sum) {
int kid, flow, rt = 0;
if (now == t) return sum;
for (int i = head[now]; i != -1 && rt < sum; i = Next[i]) {
head[now] = i;
kid = to[i];
if (lev[kid] == lev[now] + 1 && v[i]) {
flow = DFS(kid, min(sum - rt, v[i]));
if (flow) {
v[i] -= flow;
v[i^1] += flow;
rt += flow;
} else lev[kid] = -1;
}
}
return rt;
}
int dinic() {
int ans = 0;
while (BFS()) {
for (int i = 0; i <= t; i++) {
head[i] = first[i];
}
ans += DFS(s, INF);
}
return ans;
}
int find(int x) {
return f[x] == x ? x : f[x] = find(f[x]);
}
void Union(int x, int y) {
if (find(x) != find(y)) f[find(x)] = find(y);
}
void input() {
scanf("%d %d %d", &n, &m, &k);
for (int i = 0; i <= n; i++) f[i] = i;
s = 0, t = n * 2 + 2;
for (int i = 0; i < m; i++) {
scanf("%d %d", &mg[i].x, &mg[i].y);
}
int a, b;
for (int i = 0; i < k; i++) {
scanf("%d %d", &a, &b);
Union(a, b);
}
memset(vis, 0, sizeof(vis));
for (int i = 1; i <= n; i++) {
addEdge(s, i, 1);
addEdge(i + n, t, 1);
}
for (int i = 0; i < m; i++) {
int a = mg[i].x, b = mg[i].y;
for (int j = 1; j <= n; j++) {
if (find(a) == find(j) && !vis[j][b]) {
vis[j][b] = 1;
addEdge(j, b + n, 1);
}
}
}
}
int build(int x) {
for (int i = 0; i < ec; i += 2) { //更换容量,清空流量
if (to[i^1] == s) {
v[i] = x;
v[i^1] = 0;
} else if (to[i] == t) {
v[i] = x;
v[i^1] = 0;
} else {
v[i] = 1;
v[i^1] = 0;
}
}
return dinic();
}
void solve() {
int l = 0, r = n + 2;
while (l < r) {
int mid = (l + r) / 2;
if (build(mid) == mid * n) {
l = mid + 1;
} else r = mid;
}
printf("%d\n", r - 1);
}
int main() {
int T;
scanf("%d", &T);
while (T--) {
init();
input();
solve();
}
return 0;
}
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hdu 3081 Marriage Match II (二分+最大流+并查集)
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原文地址:http://blog.csdn.net/llx523113241/article/details/47449087