HDU3072(Intelligence System)

Intelligence System

Problem Description
After a day, ALPCs finally complete their ultimate intelligence system, the purpose of it is of course for ACM … …
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!

Input
There are several test cases.
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.

Output
The minimum total cost for inform everyone.
Believe kzc_tc’s working! There always is a way for him to communicate with all other intelligence personnel.

Sample Input
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

Sample Output
150
100
50

题意

给定一张有向图，从0号节点开始发送消息给各个点，如果两个点之间可以相互到达那么这两个点之间的通信费用为0，问最少需要花费多少钱。

思路

Tarjan + 缩点，最开始我是先缩点，缩完之后对边排序做kruskal形成一个最小缩点生成树，好像一看没啥问题测试数据也过了，但是一直WA。后来发现不对，最小生成树是针对无向图的，这个图是有向图。

如果是这组测试数据，kruskal + 并查集维护的答案是2，正确答案应该是 101。

最小树形图：最小树形图就是给有向带权图中指定一个特殊的点root，求一棵以root为根的有向生成树T，并且T中所有边的总权值最小。

正确姿势：Tarjan + 缩点 + 最小树形图，还是先缩点，同一个强连通分量中通信费用是0忽略。并且是从0号根节点出发的，找到最小的入边，最后去缩点边权值就好。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <stack>
#include <cmath>
#include <cstdlib>
#pragma warning(disable:4996)
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 50005;
struct edge{
	int from;
	int to;
	int w;
	int next;
}e[maxn<<1];
stack<int>s;
int head[maxn];			
int dfn[maxn];			
int low[maxn];		
int scc[maxn];			
int sum[maxn];			
int res,cnt,tot;
inline void clear_set()
{
	cnt = tot = res = 0;
	memset(head,-1,sizeof(head));
	memset(dfn,0,sizeof(dfn));
	memset(low,0,sizeof(low));
	memset(scc,0,sizeof(scc));
	memset(sum,0x3f,sizeof(sum));
	while(!s.empty())		s.pop();
}
inline void addedge(int x,int y,int z)
{
	e[tot].from = x;
	e[tot].to = y;
	e[tot].w = z;
	e[tot].next = head[x];
	head[x] = tot++;
}
inline void tarjan(int x)
{
	dfn[x] = low[x] = ++cnt;
	s.push(x);
	for(int i = head[x];~i;i = e[i].next){
		int y = e[i].to;
		if(!dfn[y]){
			tarjan(y);
			low[x] = min(low[x],low[y]);
		}
		else if(!scc[y]){
			low[x] = min(low[x],dfn[y]);
		}
	}
	if(low[x] == dfn[x]){
		res++;
		while(true){
			int t = s.top();
			s.pop();
			scc[t] = res;
			if(t == x)		break;
		}
	}
}
int main()
{
	int n,m;
	while(~scanf("%d%d",&n,&m)){
		clear_set();
		for(int i = 0;i < m;i++){
			int x,y,z;
			scanf("%d%d%d",&x,&y,&z);
			addedge(x,y,z);
		}
		for(int i = 0;i < n;i++){
			if(!dfn[i]){
				tarjan(i);
			}
		}
		for(int i = 0;i < m;i++){			
			int x = e[i].from;
			int y = e[i].to;
			if(scc[x] == scc[y])	continue;				
			if(scc[y] == scc[0])	continue;				
			sum[scc[y]] = min(sum[scc[y]],e[i].w);			//更新最小的两个缩点之间的边权值
		}
		int ans = 0;
		for(int i = 1;i <= res;i++){
			if(sum[i] != inf){				
				ans += sum[i];
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}

愿你走出半生，归来仍是少年~