# 洛谷 [P3623] 免费道路

## 有 k 条特殊边的生成树

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
using namespace std;
const int MAXN = 100005;
int init() {
int rv = 0, fh = 1;
char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') fh = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
rv = (rv<<1) + (rv<<3) + c - '0';
c = getchar();
}
return fh * rv;
}
int n, m, nume, head[MAXN], fa[MAXN], k, cnt, ans[MAXN], ind;
struct edge{
int to, nxt, opt;
}e[MAXN << 1];
void adde(int from, int to, int opt) {
e[++nume].to = to;
e[nume].opt = opt;
}
int find(int x) {
if(x != fa[x]) return fa[x] = find(fa[x]);
return fa[x];
}
void merge(int x, int y) {
int r1 = find(x), r2 = find(y);
if(r1 != r2) {
fa[r1] = r2;
}
}
void input() {
for(int i = 1; i <= n; i++) fa[i] = i;
}
bool chk() {
for(int i = 2; i <= n; i++) if(find(i) != find(1)) return 0;
return 1;
}
int main() {
n = init(); m = init(); k = init();
input();
for(int i = 1; i <= m; i++) {
int u = init(), v = init(), opt = init();
if(opt == 1) merge(u, v);
}
for(int i = 1; i <= m * 2; i += 2) {
if(e[i].opt == 0) {
int r1 = find(e[i].to), r2 = find(e[((i - 1) ^ 1) + 1].to);
if(r1 != r2) {
fa[r1] = r2;
ans[++cnt] = i;
ind++;
}
}
}
if(cnt > k || !chk()) {
printf("no solution\n");
return 0;
}
input();
for(int i = 1; i <= nume; i += 2) {
if(e[i].opt == 0) merge(e[i].to, e[((i - 1) ^ 1) + 1].to);
}
for(int i = 1; i <= m * 2; i += 2) {
if(e[i].opt == 1) {
int r1 = find(e[i].to), r2 = find(e[((i - 1) ^ 1) + 1].to);
if(r1 != r2) {
fa[r1] = r2;
ans[++cnt] = i;
}
}
}
if(!chk()) {
printf("no solution\n");
return 0;
}
input();
for(int i = 1; i <= cnt; i++) {
int t = ans[i];
merge(e[t].to, e[((t - 1) ^ 1) + 1].to);
}
for(int i = 1; i <= nume; i += 2) {
if(ind == k) break;
if(e[i].opt == 0) {
int r1 = find(e[i].to), r2 = find(e[((i - 1) ^ 1) + 1].to);
if(r1 != r2) {
fa[r1] = r2;
ans[++cnt] = i;
ind++;
}
}
}
if(ind != k) {
printf("no solution\n");
return 0;
}
for(int i = 1; i <= nume; i += 2) {
if(e[i].opt == 1) {
int r1 = find(e[i].to), r2 = find(e[((i - 1) ^ 1) + 1].to);
if(r1 != r2) {
fa[r1] = r2;
ans[++cnt] = i;
}
}
}
sort(ans + 1, ans + cnt + 1);
for(int i = 1 ; i <= cnt; i++) {
int t = ans[i];
printf("%d %d %d\n", e[((t - 1) ^ 1) + 1].to, e[t].to, e[t].opt);
}
return 0;
}
posted @ 2018-04-28 11:37  Mr_Wolfram  阅读(...)  评论(...编辑  收藏