Dinic模板
2015-10-09 21:00
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#include<iostream>
#include<cstdio>
#include<string.h>
#include<string>
#include<stack>
#include<set>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<sstream>
#include<queue>
#define ll __int64
#define lll unsigned long long
#define MAX 1000009
#define MAXN 2009
#define eps 1e-8
#define INF 0x7fffffff
#define mod 1000000007
#define clr(a) memset(a,0,sizeof(a))
#define clr1(a) memset(a,-1,sizeof(a))
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
inline ll Max(ll a,ll b)
{
return a>b?a:b;
}
inline ll Min(ll a,ll b)
{
return a<b?a:b;
}
const int Num = 500;
const int Mum = 100000;
using namespace std;
/*
题意:
想法:拆点网络最大流
主要是构图思想,每种网络流其实都是一种可行方案。
*/
struct Edge{
int from,to,flow,cap, nex; //一条边有起点,终点,最大流量,当前流量,下一个节点的指针
}edge[Mum*2];//双向边,注意RE的情况 注意这个模版是 相同起末点的边 合并流量
int head[Num],edgenum;//2个要初始化-1和0
void addedge(int u,int v,int cap){//网络流要加反向弧
Edge E={u,v,0,cap,head[u]};
edge[edgenum]=E;
head[u]=edgenum++;
Edge E2={v,u,0,0,head[v]}; //这里的cap若是单向边要为0
edge[edgenum]=E2;
head[v]=edgenum++;
}
int dis[Num],cur[Num];//距离起点的距离 cur[i]表示i点正在考虑的边 优化不再考虑已经用过的点 初始化为head
bool vis[Num];
bool BFS(int Start,int End){
memset(vis,0,sizeof(vis));
memset(dis,-1,sizeof(dis));
queue<int>Q;
while(!Q.empty())
Q.pop();
Q.push(Start); dis[Start]=0; vis[Start]=1;
while(!Q.empty())
{
int u = Q.front(); Q.pop();
for(int i=head[u];i!=-1;i=edge[i].nex){
Edge E =edge[i];
if(!vis[E.to] && E.cap>E.flow)
{
vis[E.to]=1;
dis[E.to]=dis[u]+1;
if(E.to==End)return true;
Q.push(E.to);
}
}
}
return false;
}
int DFS(int x, int a,int End){//流入x 的流量是a
if(x==End || a==0)return a;
int flow = 0, f;
for(int& i=cur[x];i!=-1;i=edge[i].nex)
{
Edge& E = edge[i];
if(dis[x]+1 == dis[E.to] && (f = DFS(E.to , Min(a, E.cap-E.flow), End))>0 )
{
E.flow += f;
edge[ i^1 ].flow -= f;//反向边要减掉
flow += f;
a -= f;
if(a==0)break;
}
}
return flow;
}
int Dinic(int Start,int End){
int flow=0;
while(BFS(Start,End)){
memcpy(cur,head,sizeof(head));//把head的数组复制过去
flow += DFS(Start, INF, End);
}
return flow;
}
void init(){
memset(head, -1, sizeof(head)); edgenum = 0;
}
int main()
{
//#ifdef ONLINE_JUDGE
//#else
// freopen("an.txt","r", stdin);
//#endif
int n,f,d;
int N;
int x,y,z;
while(~scanf("%d%d%d",&n,&f,&d))
{
init();
N = 1 + f + n + n + d + 1;
for(int i = 2; i<=f+1; i++)
{
addedge(1,i,1);
}
for(int i = 1 + f + 1; i<=1 + f + n; i++)
{
addedge(i, i + n,1);
}
for(int i = 1 + f + n + n + 1; i<=1 + f + n + n + d; i++)
{
addedge(i,N,1);
}
for(int j = 1; j<=n; j++)
{
scanf("%d%d",&x,&y);
for(int i = 0; i<x; i++)
{
scanf("%d",&z);
addedge(z + 1,j + f + 1,1);
}
for(int i = 0; i<y; i++)
{
scanf("%d",&z);
addedge( 1 + f + n + j, 1 + f + n + n + z,1);
}
}
int flow = Dinic(1,N);
printf("%d\n",flow);
}
return 0;
}
#include<cstdio>
#include<string.h>
#include<string>
#include<stack>
#include<set>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<sstream>
#include<queue>
#define ll __int64
#define lll unsigned long long
#define MAX 1000009
#define MAXN 2009
#define eps 1e-8
#define INF 0x7fffffff
#define mod 1000000007
#define clr(a) memset(a,0,sizeof(a))
#define clr1(a) memset(a,-1,sizeof(a))
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
inline ll Max(ll a,ll b)
{
return a>b?a:b;
}
inline ll Min(ll a,ll b)
{
return a<b?a:b;
}
const int Num = 500;
const int Mum = 100000;
using namespace std;
/*
题意:
想法:拆点网络最大流
主要是构图思想,每种网络流其实都是一种可行方案。
*/
struct Edge{
int from,to,flow,cap, nex; //一条边有起点,终点,最大流量,当前流量,下一个节点的指针
}edge[Mum*2];//双向边,注意RE的情况 注意这个模版是 相同起末点的边 合并流量
int head[Num],edgenum;//2个要初始化-1和0
void addedge(int u,int v,int cap){//网络流要加反向弧
Edge E={u,v,0,cap,head[u]};
edge[edgenum]=E;
head[u]=edgenum++;
Edge E2={v,u,0,0,head[v]}; //这里的cap若是单向边要为0
edge[edgenum]=E2;
head[v]=edgenum++;
}
int dis[Num],cur[Num];//距离起点的距离 cur[i]表示i点正在考虑的边 优化不再考虑已经用过的点 初始化为head
bool vis[Num];
bool BFS(int Start,int End){
memset(vis,0,sizeof(vis));
memset(dis,-1,sizeof(dis));
queue<int>Q;
while(!Q.empty())
Q.pop();
Q.push(Start); dis[Start]=0; vis[Start]=1;
while(!Q.empty())
{
int u = Q.front(); Q.pop();
for(int i=head[u];i!=-1;i=edge[i].nex){
Edge E =edge[i];
if(!vis[E.to] && E.cap>E.flow)
{
vis[E.to]=1;
dis[E.to]=dis[u]+1;
if(E.to==End)return true;
Q.push(E.to);
}
}
}
return false;
}
int DFS(int x, int a,int End){//流入x 的流量是a
if(x==End || a==0)return a;
int flow = 0, f;
for(int& i=cur[x];i!=-1;i=edge[i].nex)
{
Edge& E = edge[i];
if(dis[x]+1 == dis[E.to] && (f = DFS(E.to , Min(a, E.cap-E.flow), End))>0 )
{
E.flow += f;
edge[ i^1 ].flow -= f;//反向边要减掉
flow += f;
a -= f;
if(a==0)break;
}
}
return flow;
}
int Dinic(int Start,int End){
int flow=0;
while(BFS(Start,End)){
memcpy(cur,head,sizeof(head));//把head的数组复制过去
flow += DFS(Start, INF, End);
}
return flow;
}
void init(){
memset(head, -1, sizeof(head)); edgenum = 0;
}
int main()
{
//#ifdef ONLINE_JUDGE
//#else
// freopen("an.txt","r", stdin);
//#endif
int n,f,d;
int N;
int x,y,z;
while(~scanf("%d%d%d",&n,&f,&d))
{
init();
N = 1 + f + n + n + d + 1;
for(int i = 2; i<=f+1; i++)
{
addedge(1,i,1);
}
for(int i = 1 + f + 1; i<=1 + f + n; i++)
{
addedge(i, i + n,1);
}
for(int i = 1 + f + n + n + 1; i<=1 + f + n + n + d; i++)
{
addedge(i,N,1);
}
for(int j = 1; j<=n; j++)
{
scanf("%d%d",&x,&y);
for(int i = 0; i<x; i++)
{
scanf("%d",&z);
addedge(z + 1,j + f + 1,1);
}
for(int i = 0; i<y; i++)
{
scanf("%d",&z);
addedge( 1 + f + n + j, 1 + f + n + n + z,1);
}
}
int flow = Dinic(1,N);
printf("%d\n",flow);
}
return 0;
}
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