lca就好了，不难理解
hdu1269：
模板题
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poj2186：
只会有一个scc属于。
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poj1236：
求加最少几条边能使整个图强连通。注意scc的入度和出度挺能用的；
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hdu2767：
终于写的有点熟了，20minac
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hdu3072：
缩点求最短路径即可；
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1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<stack> 5 #include<algorithm> 6 using namespace std; 7 #define rep(i,n) for(int i=1;i<=n;i++) 8 #define clr(x,c) memset(x,c,sizeof(x)) 9 struct edge{ 10 int to,dis,next; 11 }; 12 edge e[100005]; 13 int dfn[50005],low[50005],h[50005],v[50005],cot[50005]; 14 int dfn_clock,scc_cnt,n,m; 15 stack<int>s; 16 int read(){ 17 int x=0; 18 char c=getchar(); 19 while(!isdigit(c)) c=getchar(); 20 while(isdigit(c)){ 21 x=x*10+c-'0'; 22 c=getchar(); 23 } 24 return x; 25 } 26 int tarjan(int x){ 27 dfn[x]=low[x]=++dfn_clock; 28 s.push(x); 29 for(int i=h[x];i;i=e[i].next){ 30 int y=e[i].to; 31 if(!dfn[y]){ 32 tarjan(y); 33 low[x]=min(low[x],low[y]); 34 } 35 else if(!v[y]){ 36 low[x]=min(low[x],dfn[y]); 37 } 38 } 39 if(low[x]==dfn[x]){ 40 scc_cnt++; 41 while(1){ 42 int o=s.top(); 43 s.pop(); 44 v[o]=scc_cnt; 45 if(o==x) 46 break; 47 } 48 } 49 } 50 int main(){ 51 while(scanf("%d%d",&n,&m)!=EOF){ 52 int cur=dfn_clock=scc_cnt=0; 53 clr(dfn,0);clr(low,0);clr(h,0);clr(v,0);clr(cot,0x7f); 54 while(!s.empty()){ 55 s.pop(); 56 } 57 rep(i,m){ 58 int u=read(),v=read(),o=read(); 59 e[++cur].to=v+1; 60 e[cur].dis=o; 61 e[cur].next=h[u+1]; 62 h[u+1]=cur; 63 } 64 rep(i,n) 65 if(!dfn[i]) 66 tarjan(i); 67 int ans=0; 68 rep(i,n) 69 for(int j=h[i];j;j=e[j].next) 70 if(v[i]!=v[e[j].to]){ 71 cot[v[e[j].to]]=min(cot[v[e[j].to]],e[j].dis); 72 } 73 rep(i,scc_cnt){ 74 if(cot[i]<0x3f3f3f3f) 75 ans+=cot[i]; 76 } 77 printf("%d ",ans); 78 } 79 return 0; 80 }
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