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
", ans[i].id, ans[i].typ);
}
return 0;
}