Intelligence System

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

题目大意：给了一个含有 n(0<n<=50000) 个节点的有向图，图中的两点之间的连接要付出代价的（经过的边权之和），但是如果这两个点之间相互可达，代价为 0。

问从给定的节点0向其他所有的点通信，所花费的最小代价是多少？

思路：如果这两个点之间相互可达（直接简介均可），代价为 0，即在一个SCC中的点连接的代价为0。所以首先SCC缩点新构图， 形成一个DAG图（有向无环图）。注意：一个SCC内的点相互连接是不需要花费的，但是连接两个SCC是要花费的，所以我们要在每个SCC中找到花费最小的点最为整个SCC的花费，这样我们连接所有的SCC时花费才最小。还要注意，0点所在的SCC花费为0。

#include <cstdio>
#include <cstring>
#include <algorithm>
#define maxn 50000 + 5000
#define maxm 100000 + 10000
#define INF 0x3f3f3f3f
using namespace std;
int n, m;

struct node {
    int u, v, w, next;
};

node edge[maxm];

int head[maxn], cnt;
int low[maxn], dfn[maxn];
int dfs_clock;
int Stack[maxn], top;
bool Instack[maxn];
int Belong[maxn];
int scc_clock;
int num[maxn];//记录每个缩点的花费。

void init(){
    cnt = 0;
    memset(head, -1, sizeof(head));
}

void addedge(int u, int v, int w){
    int i;
    for(i = head[u]; i != -1; i = edge[i].next){
        if(edge[i].v == v)
            break;
    }
    if(i == -1){
        edge[cnt] = {u, v, w, head[u]};
        head[u] = cnt++;
    }
    else//有重边，更新这条边最小的花费
        edge[i].w = min(edge[i].w, w);
}

void getmap(){
    int a, b, c;
    while(m--){
        scanf("%d%d%d", &a, &b, &c);
        a++, b++;
        addedge(a, b, c);
    }
}

void Tarjan(int u){
    int v;
    low[u] = dfn[u] = ++dfs_clock;
    Stack[top++] = u;
    Instack[u] = true;
    for(int i = head[u]; i != -1; i = edge[i].next){
        v = edge[i].v;
        if(!dfn[v]){
            Tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if(Instack[v])
            low[u] = min(low[u], dfn[v]);
    }
    if(dfn[u] == low[u]){
        scc_clock++;
        do{
            v = Stack[--top];
            Instack[v] = false;
            Belong[v] = scc_clock;
        }
        while(v != u);
    }
}

void find(){
    memset(low, 0, sizeof(low));
    memset(dfn, 0, sizeof(dfn));
    memset(Belong, 0, sizeof(Belong));
    memset(Stack, 0, sizeof(Stack));
    memset(Instack, false, sizeof(false));
    dfs_clock = scc_clock = top = 0;
    for(int i = 1; i <= n ; ++i){
        if(!dfn[i])
            Tarjan(i);
    }
}

void suodian(){
    for(int i = 1; i <= scc_clock; ++i)
        num[i] = INF;
    for(int i = 0; i < cnt; ++i){
        int u = Belong[edge[i].u];
        int v = Belong[edge[i].v];
        if(u != v){
            //跟新每个缩点的最小花费
            num[v] = min(num[v], edge[i].w);
        }
    }
}

void solve(){
    int ans = 0;
    //printf("%d\n", scc_clock);
    for(int i = 1; i <= scc_clock; ++i){
        //printf("%d\n", num[i]);
        if(Belong[1] != i)
            ans += num[i];
    }
    printf("%d\n", ans);
}

int main (){
    while(scanf("%d%d", &n, &m) != EOF){
        init();
        getmap();
        find();
        suodian();
        solve();
    }
    return 0;
}

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