UVA-10249 - The Grand Dinner(最大流)

题意:

N个队伍在M张桌子上吃饭,问你,是否存在一种方案,使得队伍中每个人都在不同的桌子上吃饭,若不能输出0,若能,输出1并输出解决方案

分析:最大流问题,只是这个需要输出解决方案.先建图,每个队伍Ai与每个桌子Bi建一条边,权值为1,因为每个队伍只能派一个人在这张桌子上吃饭.同时建立超级源点S和超级汇点T,S向每个队伍建一条边,权值为队伍人数.每张桌子向T建一条边,权值为桌子容量!

求最大流ans,若ans等于所有人数,则有解决方案,否则无解决方案;

对与解决方案,对与每条边,起始点要求是队伍标号,终点是桌子标号,且流量flow要与容量cap相等,说明这条边被选择了!

// File Name: 10249.cpp
// Author: Zlbing
// Created Time: 2013/4/22 16:14:33

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
using namespace std;
#define CL(x,v); memset(x,v,sizeof(x));
#define INF 0x3f3f3f3f
#define LL long long
#define REP(i,r,n) for(int i=r;i<=n;i++)
#define RREP(i,n,r) for(int i=n;i>=r;i--)

const int MAXN=200;
struct Edge{
int from,to,cap,flow;
};
bool cmp(const Edge& a,const Edge& b){
return a.from < b.from || (a.from == b.from && a.to < b.to);
}
struct Dinic{
int n,m,s,t;
vector<Edge> edges;
vector<int> G[MAXN];
bool vis[MAXN];
int d[MAXN];
int cur[MAXN];
void init(int n){
this->n=n;
for(int i=0;i<=n;i++)G[i].clear();
edges.clear();
}
void AddEdge(int from,int to,int cap){
edges.push_back((Edge){from,to,cap,0});
edges.push_back((Edge){to,from,0,0});//当是无向图时，反向边容量也是cap,有向边时，反向边容量是0
m=edges.size();
G[from].push_back(m-2);
G[to].push_back(m-1);
}
bool BFS(){
CL(vis,0);
queue<int> Q;
Q.push(s);
d[s]=0;
vis[s]=1;
while(!Q.empty()){
int x=Q.front();
Q.pop();
for(int i=0;i<G[x].size();i++){
Edge& e=edges[G[x][i]];
if(!vis[e.to]&&e.cap>e.flow){
vis[e.to]=1;
d[e.to]=d[x]+1;
Q.push(e.to);
}
}
}
return vis[t];
}
int DFS(int x,int a){
if(x==t||a==0)return a;
int flow=0,f;
for(int& i=cur[x];i<G[x].size();i++){
Edge& e=edges[G[x][i]];
if(d[x]+1==d[e.to]&&(f=DFS(e.to,min(a,e.cap-e.flow)))>0){
e.flow+=f;
edges[G[x][i]^1].flow-=f;
flow+=f;
a-=f;
if(a==0)break;
}
}
return flow;
}
//当所求流量大于need时就退出，降低时间
int Maxflow(int s,int t,int need){
this->s=s;this->t=t;
int flow=0;
while(BFS()){
CL(cur,0);
flow+=DFS(s,INF);
if(flow>need)return flow;
}
return flow;
}
//最小割割边
vector<int> Mincut(){
BFS();
vector<int> ans;
for(int i=0;i<edges.size();i++){
Edge& e=edges[i];
if(vis[e.from]&&!vis[e.to]&&e.cap>0)ans.push_back(i);
}
return ans;
}
void Reduce(){
for(int i = 0; i < edges.size(); i++) edges[i].cap -= edges[i].flow;
}
void ClearFlow(){
for(int i = 0; i < edges.size(); i++) edges[i].flow = 0;
}
};
Dinic solver;
int A[MAXN],B[MAXN];
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
if(n==0&&m==0)break;
solver.init(n+m+1);
int sum=0;
int s=0,t=n+m+1;
REP(i,1,n){
scanf("%d",&A[i]);
sum+=A[i];
}
REP(i,1,m)scanf("%d",&B[i]);
REP(i,1,n)
REP(j,1,m)
solver.AddEdge(i,n+j,1);
REP(i,1,n)
solver.AddEdge(s,i,A[i]);
REP(i,1,m)
solver.AddEdge(i+n,t,B[i]);
int ans=solver.Maxflow(s,t,INF);
if(ans==sum)
{
vector<int> C[MAXN];
for(int i=0;i<solver.edges.size();i++)
{
Edge e=solver.edges[i];
//printf("e.from--%d  e.to--%d\n",e.from,e.to);
if(e.from>=1&&e.from<=n&&e.to>n&&e.to<=n+m&&e.flow==e.cap){
C[e.from].push_back(e.to-n);
}
}
printf("1\n");
for(int i=1;i<=n;i++){
for(int j=0;j<C[i].size();j++)
{
if(!j)printf("%d",C[i][j]);
else printf(" %d",C[i][j]);
}
printf("\n");
}
}
else printf("0\n");

}
return 0;
}