标签:情况 tracking 否则 女性 ber uri cas mem together
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3
一组配对情况为全部的女性都有一个与之配对的男性(一对一的关系)。假设还有其它组配对情况,那么全部的女性配对不能够再与原来的男性配成对。问最多有多少组配对情况。
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define captype int
const int MAXN = 100010; //点的总数
const int MAXM = 4000100; //边的总数
const int INF = 1<<30;
struct EDG{
int to,next;
captype cap,flow;
}edg[MAXM];
int eid,head[MAXN];
int gap[MAXN];
int dis[MAXN];
int cur[MAXN];
int pre[MAXN];
void init(){
eid=0;
memset(head,-1,sizeof(head));
}
void addEdg(int u,int v,captype c,captype rc=0){
edg[eid].to=v; edg[eid].next=head[u];
edg[eid].cap=c; edg[eid].flow=0; head[u]=eid++;
edg[eid].to=u; edg[eid].next=head[v];
edg[eid].cap=rc; edg[eid].flow=0; head[v]=eid++;
}
captype maxFlow_sap(int s,int t,int n){
memset(gap,0,sizeof(gap));
memset(dis,0,sizeof(dis));
memcpy(cur,head,sizeof(head));
pre[s]=-1;
gap[0]=n;
captype ans=0;
int u=s;
while(dis[s]<n){
if(u==t){
captype mint=INF;
int id;
for(int i=pre[u]; i!=-1; i=pre[edg[i^1].to])
if(mint>edg[i].cap-edg[i].flow){
mint=edg[i].cap-edg[i].flow;
id=i;
}
for(int i=pre[u]; i!=-1; i=pre[edg[i^1].to]){
edg[i].flow+=mint;
edg[i^1].flow-=mint;
}
ans+=mint;
u=edg[id^1].to;
continue;
}
bool flag=0;
for(int i=cur[u]; i!=-1; i=edg[i].next)
if(edg[i].cap>edg[i].flow&&dis[u]==dis[edg[i].to]+1){
cur[u]=pre[edg[i].to]=i;
flag=true;
break;
}
if(flag){
u=edg[cur[u]].to;
continue;
}
int minh=n;
for(int i=head[u]; i!=-1; i=edg[i].next)
if(edg[i].cap>edg[i].flow && minh>dis[edg[i].to]){
cur[u]=i; minh=dis[edg[i].to];
}
gap[dis[u]]--;
if(!gap[dis[u]]) return ans;
dis[u]=minh+1;
gap[dis[u]]++;
if(u!=s)
u=edg[pre[u]^1].to;
}
return ans;
}
int fath[MAXN];
int findroot(int x){
if(x!=fath[x])
fath[x]=findroot(fath[x]);
return fath[x];
}
void setroot(int x,int y){
x=findroot(x);
y=findroot(y);
fath[x]=y;
}
void rebuildMap(int mapt[255][255],int n){//处理朋友之间的关系
int mp[255][255]={0};
for(int i=1; i<=n; i++)
fath[i]=findroot(i);
for(int i=1; i<=n; i++){
int j=fath[i];
for(int e=1; e<=n; e++)
mp[j][e]|=mapt[i][e];
}
for(int i=1; i<=n; i++){
int j=fath[i];
for(int e=1; e<=n; e++)
mapt[i][e]=mp[j][e];
}
}
int main()
{
int T,n,m,k,f,mapt[255][255];
int u,v;
scanf("%d",&T);
while(T--){
scanf("%d%d%d%d",&n,&m,&k,&f);
init();
memset(mapt,0,sizeof(mapt));
for(int i=1; i<=n; i++)
fath[i]=i;
while(m--){
scanf("%d%d",&u,&v);
mapt[u][v]=1;
}
while(f--){
scanf("%d%d",&u,&v);
setroot(u,v);
}
rebuildMap(mapt,n);
int s=0, t=3*n+1;
for(int i=1; i<=n; i++){
addEdg(s,i,0);
addEdg(i,i+n,k);
for(int j=1; j<=n; j++)
if(mapt[i][j])
addEdg(i,j+2*n,1);
else
addEdg(i+n,j+2*n,1);
addEdg(i+2*n,t,0);
}
int ans=0 , l=0 , r=n ,mid;
while(l<=r){
mid=(l+r)>>1;
for(int i=0; i<eid; i++)
edg[i].flow=0;
for(int i=head[s]; i!=-1; i=edg[i].next)
edg[i].cap=mid;
for(int i=head[t]; i!=-1; i=edg[i].next)
edg[i^1].cap=mid;
if(n*mid==maxFlow_sap(s,t,t+1))
ans=mid,l=mid+1;
else
r=mid-1;
}
printf("%d\n",ans);
}
}
HDU 3277 Marriage Match III(并查集+二分答案+最大流SAP)拆点,经典
标签:情况 tracking 否则 女性 ber uri cas mem together
原文地址:http://www.cnblogs.com/yutingliuyl/p/7400880.html