[HNOI2006]公路修建问题 BZOJ1196 Kruskal

题目描述

输入输出格式

输入格式：

在实际评测时，将只会有m-1行公路

输出格式：

输入输出样例

输入样例#1：
复制

4 2 5
1 2 6 5
1 3 3 1
2 3 9 4
2 4 6 1

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<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#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<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> 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;
}