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; }