[HNOI2006]公路修建问题 BZOJ1196 Kruskal-CFANZ编程社区

题目描述

[HNOI2006]公路修建问题 BZOJ1196 Kruskal_i++

输入输出格式

输入格式：

[HNOI2006]公路修建问题 BZOJ1196 Kruskal_#include_02

[HNOI2006]公路修建问题 BZOJ1196 Kruskal_#include_03

在实际评测时，将只会有m-1行公路

输出格式：

[HNOI2006]公路修建问题 BZOJ1196 Kruskal_i++_04

输入输出样例

输入样例#1：

复制

输出样例#1： 复制

6 
1 1 
2 1 
4 1
样例貌似有点问题；
其实就是按照贪心从小到大排序就行了；
坑点就是取的maxx要一直维护（可能出现有的2级道路花费>1级道路）

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
//#include
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair pii;

inline int rd() {
  int x = 0;
  char c = getchar();
  bool f = false;
  while (!isdigit(c)) {
    if (c == '-') f = true;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return f ? -x : x;
}


ll gcd(ll a, ll b) {
  return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }



/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
  if (!b) {
    x = 1; y = 0; return a;
  }
  ans = exgcd(b, a%b, x, y);
  ll t = x; x = y; y = t - a / b * y;
  return ans;
}
*/
int n, K, m;
struct node {
  int x, y;
  int ct1, ct2;
  int used;
  int fy;
  int id;
}e[maxn],a[maxn];
bool cmp(node a, node b) {
  return a.ct1 < b.ct1;
}
bool cmp2(node a, node b) {
  return a.ct2 < b.ct2;
}

struct o {
  int id, typ;
}ans[maxn];

int fa[maxn];
void init() {
  for (int i = 0; i <= n; i++)fa[i] = i;
}

int findfa(int x) {
  if (x == fa[x])return x;
  return fa[x] = findfa(fa[x]);
}

void merge(int u, int v) {
  int p = findfa(u);
  int q = findfa(v);
  if (p != q)fa[p] = q;
}
bool cmp3(o a, o b) {
  return a.id < b.id;
}

int main()
{
  //ios::sync_with_stdio(0);
  cin >> n >> K >> m; init();
  for (int i = 1; i < m; i++) {
    int u, v, c1, c2;
    u = rd(); v = rd(); c1 = rd(); c2 = rd();
    e[i].x = u; e[i].y = v; e[i].ct1 = c1; e[i].ct2 = c2;
    e[i].id = i;
  }
  sort(e + 1, e + m, cmp);
  int tot = 0;
  int maxx = -inf;
  for (int i = 1; i < m; i++) {
    int u = e[i].x;
    int v = e[i].y;
    if (findfa(u) != findfa(v)) {
      e[i].used = 1; merge(u, v);
      maxx = max(maxx, e[i].ct1);
      ans[++tot].id = e[i].id; ans[tot].typ = 1;
      if (tot >= K)break;
    }
  }
  sort(e + 1, e + m, cmp2);
  for (int i = 1; i < m; i++) {
    if (!e[i].used) {
      int u = e[i].x;
      int v = e[i].y;
      if (findfa(u) != findfa(v)) {
        merge(u, v); ans[++tot].id = e[i].id;
        ans[tot].typ = 2;
        e[i].used = 1; maxx = max(maxx, e[i].ct2);
        if (tot >= n - 1)break;
      }
    }
  }
  sort(ans + 1, ans + 1 + tot, cmp3);
  cout << maxx << endl;
  for (int i = 1; i <= tot; i++) {
    printf("%d %d\n", ans[i].id, ans[i].typ);
  }
  
  return 0;
}

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