HDU 4738 割边模板

HDU 4738

  • 裸求割边的存在一起割边中的最小值
  1. 割边不存在输出-1
  2. 图不是连通图,则已经达到题目条件.士兵花费为0
  3. 坑点: 割边存在最小值为0,需要一个士兵抗炸弹
  • 割边模板
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#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn = 1e3+10;
const int maxm = 1e6+10;

int head[maxn],ver[maxm],next1[maxm],dist[maxm];
int dfn[maxn],low[maxn],n,m,tot,num;
//bool bridge[maxm];
int ans;

void add(int x,int y,int v) {
    ver[++tot] = y;
    dist[tot] = v;
    next1[tot] = head[x];
    head[x] = tot;
}

void tarjan(int x,int in_edge) {
    dfn[x] = low[x] = ++num;
    for (int i=head[x]; i; i=next1[i]) {
        int to = ver[i];
        if (!dfn[to]) {
            tarjan(to,i);
            low[x] = min(low[x],low[to]);
            if (low[to] > dfn[x]) {
                //cout << "??" << endl;
                //bridge[i] = bridge[i^1] = 1;
                ans = min(ans,dist[i]);
                ans = min(ans,dist[i^1]);
            }
        } else if (i != (in_edge ^ 1))
            low[x] = min(low[x],dfn[to]);
    }
}

void init() {
    ans = 0x3f3f3f3f;
    memset(head,0,sizeof head);
    //memset(ver,0,sizeof ver);
    memset(next1,0,sizeof next1);
    memset(dist,0x3f,sizeof dist);
    memset(dfn,0,sizeof dfn);
    memset(low,0,sizeof low);
    tot = 1;
    num = 0;
}

int main() {
    freopen("1.in","r",stdin);
    while (scanf("%d%d",&n,&m) && (n+m)) {
        init();
        for (int i=1; i<=m; ++i) {
            int x,y,v; cin >> x >> y >> v;
            add(x,y,v);
            add(y,x,v);
        }
        int flag = 0;
        for (int i=1; i<=n; ++i)
            if (!dfn[i]) {
                flag++;
                tarjan(i,0);
            }
        if (ans == 0x3f3f3f3f) { cout << "-1" << endl; }
        else if (flag >= 2) cout << "0" << endl;
        else if (ans == 0) cout << "1" << endl;
        else cout << ans << endl;
    }
    return 0;
}

算法

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