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题目要的并不是最大匹配下得到的最大的结果。
网上流行的做法是加边。其实,在连续增广的时候求得一个可行流的总费用为负就停止这样就行了。
1 #include<cstdio> 2 #include<cstring> 3 #include<queue> 4 #include<algorithm> 5 using namespace std; 6 #define INF (1<<30) 7 #define MAXN 222 8 #define MAXM 22222 9 struct Edge{ 10 int u,v,cap,cost,next; 11 }edge[MAXM]; 12 int head[MAXN]; 13 int NV,NE,vs,vt; 14 15 void addEdge(int u,int v,int cap,int cost){ 16 edge[NE].u=u; edge[NE].v=v; edge[NE].cap=cap; edge[NE].cost=cost; 17 edge[NE].next=head[u]; head[u]=NE++; 18 edge[NE].u=v; edge[NE].v=u; edge[NE].cap=0; edge[NE].cost=-cost; 19 edge[NE].next=head[v]; head[v]=NE++; 20 } 21 bool vis[MAXN]; 22 int d[MAXN],pre[MAXN]; 23 bool SPFA(){ 24 for(int i=0;i<NV;++i){ 25 vis[i]=0; 26 d[i]=INF; 27 } 28 vis[vs]=1; 29 d[vs]=0; 30 queue<int> que; 31 que.push(vs); 32 while(!que.empty()){ 33 int u=que.front(); que.pop(); 34 for(int i=head[u]; i!=-1; i=edge[i].next){ 35 int v=edge[i].v; 36 if(edge[i].cap && d[v]>d[u]+edge[i].cost){ 37 d[v]=d[u]+edge[i].cost; 38 pre[v]=i; 39 if(!vis[v]){ 40 vis[v]=1; 41 que.push(v); 42 } 43 } 44 } 45 vis[u]=0; 46 } 47 return d[vt]!=INF; 48 } 49 int MCMF(){ 50 int res=0; 51 while(SPFA()){ 52 int flow=INF,cost=0; 53 for(int u=vt; u!=vs; u=edge[pre[u]].u){ 54 flow=min(flow,edge[pre[u]].cap); 55 } 56 for(int u=vt; u!=vs; u=edge[pre[u]].u){ 57 edge[pre[u]].cap-=flow; 58 edge[pre[u]^1].cap+=flow; 59 cost+=flow*edge[pre[u]].cost; 60 } 61 if(cost>=0) break; 62 res+=cost; 63 } 64 return res; 65 } 66 int main(){ 67 int n,a[111],b; 68 while(~scanf("%d",&n)&&n){ 69 for(int i=1;i<=n;++i) scanf("%d",a+i); 70 vs=0; vt=n<<1|1; NV=vt+1; NE=0; 71 memset(head,-1,sizeof(head)); 72 for(int i=1;i<=n;++i){ 73 addEdge(vs,i,1,0); 74 addEdge(i+n,vt,1,0); 75 } 76 for(int i=1;i<=n;++i){ 77 for(int j=1;j<=n;++j){ 78 scanf("%1d",&b); 79 if(b) addEdge(i,j+n,1,-(a[i]^a[j])); 80 } 81 } 82 printf("%d\n",-MCMF()); 83 } 84 return 0; 85 }
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原文地址:http://www.cnblogs.com/WABoss/p/4852052.html