模板[有向图的强连通分量] -hdu 3072
三重算法：
一、 Kosaraju算法
二、 Trajan算法
三、 Gabow算法
前两种马马虎虎刚学会把模板套上去
对应题目：HDU 3072
给你n个点，m条边，每条边有一个权值（传送message代价），已知强连通分支内部不需花费，求minimal cost
步骤：
1.缩点n->scc个点
2.将到这scc个点的最小代价计算出来
3.相加
Trajan模板：
#include<iostream> #include<algorithm> using namespace std; #define min(x,y) x>y?y:x #define maxn 50005 #define max(x,y) x>y?x:y #define INF 10000000 int low[maxn],dfn[maxn],stack[maxn],Belong[maxn]; bool instack[maxn]; int n,stop,scc,step,cost[maxn]; struct Node { int to; int val; struct Node* next; }Edge[maxn],g_Temp[maxn*2]; int g_Pos = 0; void Tarjan(int v) { dfn[v]=low[v] = ++step; instack[v] = true; stack[++stop] = v; Node *temp = Edge[v].next; int t; while(temp!=NULL) { t = temp->to; if(!dfn[t]) { Tarjan(t); low[v] = min(low[v],low[t]); } else if(instack[t] && low[v] > dfn[t]) low[v] = dfn[t]; temp = temp->next; } if(dfn[v] == low[v]) { scc++; do{ t = stack[stop--]; instack[t] = false; Belong[t] = scc; }while(t!=v); } } void solve(int n) { int i; memset(dfn,0,sizeof(dfn)); memset(instack,false,sizeof(instack)); stop = scc = step = 0; for( i = 1; i <= n; i ++) { if(!dfn[i]) { Tarjan(i); } } } void Init() { g_Pos = 0; for(int i = 0 ; i <= n; i ++) { Edge[i].next = NULL; } } void add_edge(int u,int v,int val) { Node *temp = &g_Temp[g_Pos++]; temp->to = v; temp->val = val; temp->next = Edge[u].next; Edge[u].next = temp; } //注意点从1开始标记 int main() { int i,j,m; while(scanf("%d%d",&n,&m)!=EOF) { Init(); int a,b,c; for(i=1;i<=m;i++) { scanf("%d%d%d",&a,&b,&c); add_edge(a+1,b+1,c); } solve(n); for(i=1;i<=scc;i++) cost[i]=INF;//cost[i] record the minimal cost of sending message (from any point) to i for(i=1;i<=n;i++) { Node *temp=Edge[i].next; while(temp!=NULL) { int v=temp->to; int w=temp->val; if(Belong[i]!=Belong[v])//don't belong to one set cost[Belong[v]]=min(cost[Belong[v]],w); temp=temp->next; } } int ans=0; for(i=1;i<=scc;i++) { if(cost[i]==INF)//the vertex whose in-degree=0 -> the root vertex(the only one who know message) continue; ans+=cost[i]; } printf("%d/n",ans); } return 0; }
Kosaraju模板：
（摘自 某大牛 的博客）
#include<stdio.h> #include<string.h> #include<vector> using namespace std; const int N=50010; const int inf=1000000000; int n, m; struct Edge { int v;//next int w;//value }e; vector<Edge> beg[N]; vector<Edge> rebeg[N]; void init() { for(int i=0;i<n;i++) { beg[i].clear(); rebeg[i].clear(); } } void add(int a, int b, int w) { e.v=b; e.w=w; beg[a].push_back(e); e.v = a; rebeg[b].push_back(e); } int order[N]; int id[N],dis[N]; int cnt, scc; void dfs(int u)//先用对原图G进行深搜形成森林(树) { for(int i=0;i<beg[u].size();i++) { int v = beg[u][i].v; if (!id[v]) { id[v]=1; dfs(v); } } order[++cnt]=u; } void redfs(int u) { for(int i=0;i<rebeg[u].size();i++) { int v=rebeg[u][i].v; if(!id[v]) { id[v]=scc; redfs(v); } } } void kosaraju(int n) { cnt=0; memset(order,0,sizeof (order)); //first dfs memset(id,0,sizeof (id)); for(int i=0;i<n;i++) { if(!id[i]) { id[i]=1; dfs(i); } } //second dfs memset(id,0,sizeof (id)); scc=0; for(int i=cnt;i>0;i--) { int u=order[i]; if(!id[u]) { id[u]=++scc; redfs(u); } } } int main() { while(scanf("%d%d",&n,&m)!=EOF) { init(); for(int i=1;i<=m;i++) { int a,b,w; scanf("%d%d%d",&a,&b,&w);//a->b,权值为w add(a,b,w); } kosaraju(n); for (int i=1;i<=scc;i++) dis[i]=inf; for(int u=0;u<n;u++) { for(int j=0;j<beg[u].size();j++) { int v=beg[u][j].v; int w=beg[u][j].w; if(id[u]!=id[v]) dis[id[v]]=min(dis[id[v]],w); } } int ans=0; for(int i=1;i<=scc;i++) { if(dis[i]==inf) dis[i]=0; ans+=dis[i]; } printf("%d/n",ans); } return 0; }
Intelligence System
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 241 Accepted Submission(s): 109
Now, kzc_tc, the head of the Intelligence Department (his code is once 48, but now 0), is sudden obtaining important information from one Intelligence personnel. That relates to the strategic direction and future development of the situation of ALPC. So it need for emergency notification to all Intelligence personnel, he decides to use the intelligence system (kzc_tc inform one, and the one inform other one or more, and so on. Finally the information is known to all).
We know this is a dangerous work. Each transmission of the information can only be made through a fixed approach, from a fixed person to another fixed, and cannot be exchanged, but between two persons may have more than one way for transferring. Each act of the transmission cost Ci (1 <= Ci <= 100000), the total cost of the transmission if inform some ones in our ALPC intelligence agency is their costs sum.
Something good, if two people can inform each other, directly or indirectly through someone else, then they belong to the same branch (kzc_tc is in one branch, too!). This case, it’s very easy to inform each other, so that the cost between persons in the same branch will be ignored. The number of branch in intelligence agency is no more than one hundred.
As a result of the current tensions of ALPC’s funds, kzc_tc now has all relationships in his Intelligence system, and he want to write a program to achieve the minimum cost to ensure that everyone knows this intelligence.
It's really annoying!
In each case, the first line is an Integer N (0< N <= 50000), the number of the intelligence personnel including kzc_tc. Their code is numbered from 0 to N-1. And then M (0<= M <= 100000), the number of the transmission approach.
The next M lines, each line contains three integers, X, Y and C means person X transfer information to person Y cost C.
Believe kzc_tc’s working! There always is a way for him to communicate with all other intelligence personnel.
3 3 0 1 100 1 2 50 0 2 100 3 3 0 1 100 1 2 50 2 1 100 2 2 0 1 50 0 1 100
150 100 50