建图:
i点拆成ai,bi两个点。源点到ai连a[i]的流量,ai到bi连INF(表示不动的士兵)bi到汇点连b[i]的流量。对于x到y的每条边连ax到by,ay到bx,流量为INF的边(士兵移动)。跑出最大流看是否等于a[i]或b[i]之和。
重点是输出方案,当最大流跑完后遍历1到N的点(表示初始状态的点)的邻接表,如果有流量flow大于0的边,就说明这个方案中沿着这条边有flow个士兵移动了。
代码:
#include <iostream> #include <cstdio> #define LL long long #include <cstring> using namespace std; #include <map> #include <queue> #include <algorithm> #include <vector> #include <queue> #include <stack> #define MAXN 205 #define INF 100000000 struct Edge{ int from,to,cap,flow; Edge(int f,int t,int c,int fl){ from=f; to=t; cap=c; flow=fl; } }; struct Dinic{ int n,m,s,t; //结点数,边数(包括反向弧),源点编号和汇点编号 vector<Edge> edges; //边表。edge[e]和edge[e^1]互为反向弧 vector<int> G[MAXN]; //邻接表,G[i][j]表示节点i和第j条边在e数组中的序号 bool vis[MAXN]; //BFS使用 int d[MAXN]; //从起点到i的距离 int cur[MAXN]; //当前弧下标 void clear_all(int n){ for(int i=0;i<n;i++) G[i].clear(); edges.clear(); } void clear_flow(){ int len=edges.size(); for(int i=0;i<len;i++) edges[i].flow=0; } void add_edge(int from,int to,int cap){ edges.push_back(Edge(from,to,cap,0)); edges.push_back(Edge(to,from,0,0)); m=edges.size(); G[from].push_back(m-2); G[to].push_back(m-1); } bool BFS(){ memset(vis,0,sizeof(vis)); queue<int> q; q.push(s); d[s]=0; vis[s]=1; while(!q.empty()){ int x=q.front(); q.pop(); int len=G[x].size(); for(int i=0;i<len;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,len=G[x].size(); for(int& i=cur[x];i<len;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; } int maxflow(int s,int t){ this->s=s; this->t=t; int flow=0; while(BFS()){ memset(cur,0,sizeof(cur)); flow+=DFS(s,INF); } return flow; } int mincut(){ //call this after maxflow int ans=0; int len=edges.size(); for(int i=0;i<len;i++){ Edge& e=edges[i]; if(vis[e.from]&&!vis[e.to]&&e.cap>0) ans++; } return ans; } void reduce(){ int len=edges.size(); for(int i=0;i<len;i++) edges[i].cap-=edges[i].flow; } }solver; int N,M; int a[MAXN]; int b[MAXN]; int mp[MAXN][MAXN]; int main(){ scanf("%d%d",&N,&M); solver.clear_all(N+1); int sum1=0,sum2=0; for(int i=1;i<=N;i++){ scanf("%d",&a[i]); sum1+=a[i]; solver.add_edge(0,i,a[i]); solver.add_edge(i,i+N,INF); } for(int i=1;i<=N;i++){ scanf("%d",&b[i]); sum2+=b[i]; solver.add_edge(i+N,N*2+1,b[i]); } if(sum1!=sum2) { printf("NO\n"); return 0; } for(int i=0;i<M;i++){ int a,b; scanf("%d%d",&a,&b); solver.add_edge(a,b+N,INF); solver.add_edge(b,a+N,INF); } int res=0; res=solver.maxflow(0,N*2+1); if(res!=sum1){ printf("NO\n"); } else { printf("YES\n"); for(int i=1;i<=N;i++){ for(int j=0;j<solver.G[i].size();j++){ Edge e=solver.edges[solver.G[i][j]]; if(e.flow){ mp[i][e.to-N]=e.flow; } } } for(int i=1;i<=N;i++){ for(int j=1;j<=N;j++){ printf("%d",mp[i][j]); if(j==N) printf("\n"); else printf(" "); } } } return 0; }