HDU 4005 The war

The war

Time Limit: 1000ms

Memory Limit: 65768KB

This problem will be judged on HDU. Original ID: 4005
64-bit integer IO format: %I64d      Java class name: Main

 

In the war, the intelligence about the enemy is very important. Now, our troop has mastered the situation of the enemy's war zones, and known that these war zones can communicate to each other directly or indirectly through the network. We also know the enemy is going to build a new communication line to strengthen their communication network. Our task is to destroy their communication network, so that some of their war zones can't communicate. Each line has its "cost of destroy". If we want to destroy a line, we must spend the "cost of destroy" of this line. We want to finish this task using the least cost, but our enemy is very clever. Now, we know the network they have already built, but we know nothing about the new line which our enemy is going to build. In this condition, your task is to find the minimum cost that no matter where our enemy builds the new line, you can destroy it using the fixed money. Please give the minimum cost. For efficiency, we can only destroy one communication line.

 

Input

The input contains several cases. For each cases, the first line contains two positive integers n, m (1<=n<=10000, 0<=m<=100000) standing for the number of the enemy's war zones (numbered from 1 to n), and the number of lines that our enemy has already build. Then m lines follow. For each line there are three positive integer a, b, c (1<=a, b<=n, 1<=c<=100000), meaning between war zone A and war zone B there is a communication line with the "cost of destroy " c.

 

Output

For each case, if the task can be finished output the minimum cost, or output ‐1.

 

Sample Input

3 2
1 2 1
2 3 2
4 3
1 2 1
1 3 2
1 4 3

Sample Output

-1
3

Hint

For the second sample input: our enemy may build line 2 to 3, 2 to 4, 3 to 4. If they build line 2 to 3, we will destroy line 1 to 4, cost 3. If they build line 2 to 4, we will destroy line 1 to 3, cost 2. If they build line 3 to 4, we will destroy line 1 to 2, cost 1. So, if we want to make sure that we can destroy successfully, the minimum cost is 3.

Source

The 36th ACM/ICPC Asia Regional Dalian Site —— Online Contest

 

解题：边双连通分量+dp

#include <bits/stdc++.h>
using namespace std;
const int maxn = 10010;
const int INF = 0x3f3f3f3f;
struct arc{
    int to,w,next;
    arc(int x = 0,int y = 0,int z = -1){
        to = x;
        w = y;
        next = z;
    }
}e[500000];
int hd[maxn],hd2[maxn],low[maxn],dfn[maxn],belong[maxn],clk,bcc,tot;
void add(int *head,int u,int v,int w){
    e[tot] = arc(v,w,head[u]);
    head[u] = tot++;
}
stack<int>stk;
void tarjan(int u,int fa){
    dfn[u] = low[u] = ++clk;
    stk.push(u);
    bool flag = false;
    for(int i = hd[u]; ~i; i = e[i].next){
        if(e[i].to == fa && !flag){
            flag = true;
            continue;
        }
        if(!dfn[e[i].to]){
            tarjan(e[i].to,u);
            low[u] = min(low[u],low[e[i].to]);
        }else low[u] = min(low[u],dfn[e[i].to]);
    }
    if(low[u] == dfn[u]){
        int v;
        ++bcc;
        do{
          v = stk.top();
          stk.pop();
          belong[v] = bcc;
        }while(v != u);
    }
}

int ret = INF,dp[maxn];
void dfs(int u,int fa){
    for(int i = hd2[u]; ~i; i = e[i].next){
        if(e[i].to == fa) continue;
        dfs(e[i].to,u);
        dp[e[i].to] = min(dp[e[i].to],e[i].w);
        if(dp[u] > dp[e[i].to]){
            ret = min(ret,dp[u]);
            dp[u] = dp[e[i].to];
        }else ret = min(ret,dp[e[i].to]);
    }
}
void init(){
    for(int i = 0; i < maxn; ++i){
        hd[i] = hd2[i] = -1;
        dfn[i] = belong[i] = 0;
        dp[i] = INF;
    }
    tot = bcc = clk = 0;
    while(!stk.empty()) stk.pop();
}
int main(){
    int n,m,u,v,w;
    while(~scanf("%d%d",&n,&m)){
        init();
        for(int i = 0; i < m; ++i){
            scanf("%d%d%d",&u,&v,&w);
            add(hd,u,v,w);
            add(hd,v,u,w);
        }
        for(int i = 1; i <= n; ++i)
            if(!dfn[i]) tarjan(i,-1);
        if(bcc == 1){
            puts("-1");
            continue;
        }
        int minW = INF,id = 0;
        for(int tt = tot, i = 0; i < tt; i += 2){
            if(belong[e[i].to] == belong[e[i^1].to]) continue;
            add(hd2,belong[e[i].to],belong[e[i^1].to],e[i].w);
            add(hd2,belong[e[i^1].to],belong[e[i].to],e[i].w);
            if(e[i].w < minW) minW = e[id = i].w;
        }
        ret = INF;
        dfs(belong[e[id].to],belong[e[id^1].to]);
        dfs(belong[e[id^1].to],belong[e[id].to]);
        printf("%d
",ret == INF?-1:ret);
    }
    return 0;
}