Marriage Match II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit:
32768/32768 K (Java/Others)
Problem 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<m<n*n,0<=f<n). n means there are 2*n children, n girls(number from 1 to n) and n boys(number from 1 to n).
Then m lines follow. Each line contains two numbers a and b, means girl a and boy b had never quarreled with each other.
Then f lines follow. Each line contains two numbers c and d, means girl c and girl d are good friends.
Output
For each case, output a number in one line. The maximal number of Marriage Match the children can play.
Sample Input
1 4 5 2 1 1 2 3 3 2 4 2 4 4 1 4 2 3
Sample Output
2
WA了一个上午，终于AC了，原来是建图时建错了。
题意：n个男孩和n个女孩，女孩选男孩，每轮游戏都必须选择不同的男孩。每轮游戏中，每个女孩可以选择和自己没有争吵的男孩，也可以选择和自己的好朋友没有争吵过的男孩。问最多可以进行几轮游戏。
分析：女孩1-n编号，男孩（n+1）- n*2编号，添加超级源点0和超级汇点2*n+1，二分枚举可以进行的轮数mid，女孩向可以选择的男孩建容量为1的边，源点向所有女孩建容量为mid的边，所有男孩向汇点建容量为mid的边。如果源点到汇点的最大流为mid*n，说明可以进行mid轮游戏，更新mid直至二分结束。处理的过程中，可以用并查集把所有互相是朋友的女生处理为一个集合。
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;
const int inf = 0x3fffffff;
const int NN = 110;
int father[NN];
int Map[NN][NN];
int src, sink, n, m, f;
struct bg_friend
{
int b, g;
}bg[10005];
template <int N, int M>
struct Isap
{
int top;
int d[N], pre[N], cur[N], gap[N];
struct Vertex{
int head;
} V[N];
struct Edge{
int v, next;
int c, f;
} E[M];
void init(){
memset(V, -1, sizeof(V));
top = 0;
}
void add_edge(int u, int v, int c){
E[top].v = v;
E[top].c = c;
E[top].f = 0;
E[top].next = V[u].head;
V[u].head = top++;
}
void add(int u,int v, int c){
add_edge(u, v, c);
add_edge(v, u, 0);
}
void set_d(int t){
queue<int> Q;
memset(d, -1, sizeof(d));
memset(gap, 0, sizeof(gap));
d[t] = 0;
Q.push(t);
while(!Q.empty()) {
int v = Q.front(); Q.pop();
++gap[d[v]];
for(int i = V[v].head; ~i; i = E[i].next) {
int u = E[i].v;
if(d[u] == -1) {
d[u] = d[v] + 1;
Q.push(u);
}
}
}
}
int sap(int s, int t, int num) {
set_d(t);
int ans = 0, u = s;
int flow = inf;
memcpy(cur, V, sizeof(V));
while(d[s] < num) {
int &i = cur[u];
for(; ~i; i = E[i].next) {
int v = E[i].v;
if(E[i].c > E[i].f && d[u] == d[v] + 1) {
u = v;
pre[v] = i;
flow = min(flow, E[i].c - E[i].f);
if(u == t) {
while(u != s) {
int j = pre[u];
E[j].f += flow;
E[j^1].f -= flow;
u = E[j^1].v;
}
ans += flow;
flow = inf;
}
break;
}
}
if(i == -1) {
if(--gap[d[u]] == 0)
break;
int dmin = num - 1;
cur[u] = V[u].head;
for(int j = V[u].head; ~j; j = E[j].next)
if(E[j].c > E[j].f)
dmin = min(dmin, d[E[j].v]);
d[u] = dmin + 1;
++gap[d[u]];
if(u != s)
u = E[pre[u] ^ 1].v;
}
}
return ans;
}
};
Isap<1000, 1000000> Sap;
struct Union_Find
{
void Init(int x)
{
for(int i = 0; i <= x; i++)
father[i] = i;
}
int Find(int x)
{
if(x != father[x])
father[x] = Find(father[x]);
return father[x];
}
void Union(int a, int b)
{
int p = Find(a);
int q = Find(b);
if(p > q)
father[p] = q;
else
father[q] = p;
}
} UF;
void build(int mid) //建边
{
memset(Map, 0, sizeof(Map));
Sap.init();
src = 0; //源点
sink = 2 * n + 1; //汇点
for(int i = 1; i <= n; i++)
{
Sap.add(src, i, mid);
Sap.add(n+i, sink, mid);
}
for(int i = 0; i < m; i++)
{
int u = bg[i].g, v = bg[i].b;
for(int j = 1; j <= n; j++)
{
if(UF.Find(j) == UF.Find(u))
{
if(!Map[j][v])
{
Map[j][v] = 1;
Sap.add(j, v+n, 1);
}
}
}
}
}
int main()
{
int i, j, T, a, b;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&n,&m,&f);
UF.Init(n);
for(i = 0; i < m; i++)
{
scanf("%d%d",&bg[i].g,&bg[i].b);
}
for(i = 0; i < f; i++)
{
scanf("%d%d",&a, &b);
UF.Union(a, b);
}
int l = 0, r = n, mid, ans = 0;
while(l <= r)
{
mid = (l + r) / 2;
build(mid);
int res = Sap.sap(src, sink, m + f + n*2);
if(res == mid * n)
{
l = mid + 1;
ans = mid;
}
else
r = mid - 1;
}
printf("%d\n", ans);
}
return 0;
}
在建图时也可以用Floyd传递女生的朋友关系。
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;
const int inf = 0x3fffffff;
const int NN = 110;
int Map[NN][NN], fri[NN][NN];
int src, sink, n, m, f;
template <int N, int M>
struct Isap
{
int top;
int d[N], pre[N], cur[N], gap[N];
struct Vertex{
int head;
} V[N];
struct Edge{
int v, next;
int c, f;
} E[M];
void init(){
memset(V, -1, sizeof(V));
top = 0;
}
void add_edge(int u, int v, int c){
E[top].v = v;
E[top].c = c;
E[top].f = 0;
E[top].next = V[u].head;
V[u].head = top++;
}
void add(int u,int v, int c){
add_edge(u, v, c);
add_edge(v, u, 0);
}
void set_d(int t){
queue<int> Q;
memset(d, -1, sizeof(d));
memset(gap, 0, sizeof(gap));
d[t] = 0;
Q.push(t);
while(!Q.empty()) {
int v = Q.front(); Q.pop();
++gap[d[v]];
for(int i = V[v].head; ~i; i = E[i].next) {
int u = E[i].v;
if(d[u] == -1) {
d[u] = d[v] + 1;
Q.push(u);
}
}
}
}
int sap(int s, int t, int num) {
set_d(t);
int ans = 0, u = s;
int flow = inf;
memcpy(cur, V, sizeof(V));
while(d[s] < num) {
int &i = cur[u];
for(; ~i; i = E[i].next) {
int v = E[i].v;
if(E[i].c > E[i].f && d[u] == d[v] + 1) {
u = v;
pre[v] = i;
flow = min(flow, E[i].c - E[i].f);
if(u == t) {
while(u != s) {
int j = pre[u];
E[j].f += flow;
E[j^1].f -= flow;
u = E[j^1].v;
}
ans += flow;
flow = inf;
}
break;
}
}
if(i == -1) {
if(--gap[d[u]] == 0)
break;
int dmin = num - 1;
cur[u] = V[u].head;
for(int j = V[u].head; ~j; j = E[j].next)
if(E[j].c > E[j].f)
dmin = min(dmin, d[E[j].v]);
d[u] = dmin + 1;
++gap[d[u]];
if(u != s)
u = E[pre[u] ^ 1].v;
}
}
return ans;
}
};
Isap<1000, 1000000> Sap;
void Floyd()
{
int i, j, k;
for(k = 1; k <= n; k++)
{
for(i = 1; i <= n; i++)
{
if(fri[i][k])
{
for(j = 1; j <= n; j++)
if(fri[k][j])
fri[i][j] = 1;
}
}
}
for(i = 1; i <= n; i++)
{
for(j = 1; j <= n; j++)
{
if(fri[i][j])
{
for(k = 1; k <= n; k++)
if(Map[j][k])
Map[i][k] = 1;
}
}
}
}
void build(int mid) //建边
{
Sap.init();
src = 0; //源点
sink = 2 * n + 1; //汇点
for(int i = 1; i <= n; i++)
{
Sap.add(src, i, mid);
Sap.add(n+i, sink, mid);
}
for(int i = 1; i <= n; i++)
{
for(int j = 1; j <= n; j++)
{
if(Map[i][j])
Sap.add(i, j+n, 1);
}
}
}
int main()
{
int i, j, T, a, b;
scanf("%d",&T);
while(T--)
{
memset(Map, 0, sizeof(Map));
memset(fri, 0, sizeof(fri));
scanf("%d%d%d",&n,&m,&f);
for(i = 0; i < m; i++)
{
scanf("%d%d",&a,&b);
Map[a][b] = 1;
}
for(i = 0; i < f; i++)
{
scanf("%d%d",&a, &b);
fri[a][b] = fri[b][a] = 1;
}
Floyd();
int l = 0, r = n, mid, ans = 0;
while(l <= r)
{
mid = (l + r) / 2;
build(mid);
int res = Sap.sap(src, sink, m + f + n*2);
if(res == mid * n)
{
l = mid + 1;
ans = mid;
}
else
r = mid - 1;
}
printf("%d\n", ans);
}
return 0;
}
hdu 3081 Marriage Match II（最大流 + 二分 + 并查集）