先scc缩点，变成DAG，显然ans=入度为0的scc个数，每个scc的答案就是scc内点权最小的值。
1 #include<bits/stdc++.h> 2 #define INF 0x7fffffff 3 using namespace std; 4 const int MAXN = 2010;//点数 5 const int MAXM = 2100;//边数 6 struct Edge { 7 int to,next; 8 }edge[MAXM]; 9 int head[MAXN],tot,a,b; 10 int Low[MAXN],DFN[MAXN],Stack[MAXN],Belong[MAXN];//Belong数组的值是1~scc 11 int Index,top; int scc;//强连通分量的个数 12 int in[MAXN],out[MAXN],val[MAXN],d[MAXN]; 13 bool Instack[MAXN]; 14 int num[MAXN];//各个强连通分量包含点的个数，数组编号1~scc //num数组不一定需要，结合实际情况 15 int cnt=0,ans=0; 16 void addedge(int u,int v){ 17 edge[tot].to = v; 18 edge[tot].next = head[u]; 19 head[u] = tot++; 20 } 21 void Tarjan(int u){ 22 int v; 23 Low[u] = DFN[u] = ++Index; 24 Stack[top++] = u; 25 Instack[u] = true; 26 for(int i = head[u];i != -1;i = edge[i].next) { 27 v = edge[i].to; 28 if( !DFN[v] ){ 29 Tarjan(v); 30 if( Low[u] > Low[v] ) 31 Low[u] = Low[v]; 32 } 33 else if(Instack[v] && Low[u] > DFN[v]) 34 Low[u] = DFN[v]; 35 } 36 if(Low[u] == DFN[u]){ 37 scc++; 38 do{ 39 v = Stack[--top]; 40 Instack[v] = false; 41 Belong[v] = scc; 42 num[scc]++; 43 }while( v != u); 44 } 45 } 46 void find_degree(int u){ 47 int v; 48 for(int i = head[u];i != -1;i = edge[i].next){ 49 v = edge[i].to; 50 if(Belong[u]==Belong[v]) continue; 51 in[Belong[v]]++; 52 //out[Belong[u]]++; 53 } 54 } 55 void solve(int N){ 56 memset(DFN,0,sizeof(DFN)); 57 memset(Instack,false,sizeof(Instack)); 58 memset(num,0,sizeof(num)); 59 Index = scc = top = 0; 60 for(int i=1;i<=N;i++) val[i]=INF; 61 for(int i = 1;i <= N;i++) 62 if(!DFN[i]) 63 Tarjan(i); 64 for(int i=1;i<=N;i++){ 65 val[Belong[i]]=min(val[Belong[i]],d[i]); 66 find_degree(i); 67 } 68 for(int i=1;i<=scc;i++){ 69 if(in[i]==0) cnt++,ans+=val[i]; 70 } 71 } 72 void init(){ 73 tot = 0; 74 cnt=0,ans=0; 75 memset(head,-1,sizeof(head)); 76 memset(in,0,sizeof(in)); 77 memset(out,0,sizeof(out)); 78 } 79 int main(){ 80 int n,m,u,v,T; 81 while(scanf("%d%d",&n,&m)!=EOF){ 82 init(); 83 for(int i=1;i<=n;i++) scanf("%d",&d[i]); 84 for(int i=1;i<=m;i++){ 85 scanf("%d%d",&u,&v); 86 addedge(u,v); 87 } 88 solve(n); 89 cout<<cnt<<' '<<ans<<endl; 90 } 91 return 0; 92 }