就是贪心的想法。。。先取1的再取0的边
验证一下满不满足要求以后，再做一遍生成树。
反正Kruskal是O(m)的无压力2333
1 /************************************************************** 2 Problem: 3624 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:60 ms 7 Memory:16416 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 12 using namespace std; 13 const int N = 20005; 14 const int M = 100005; 15 const int Maxlen = 15000010; 16 17 struct edges { 18 int x, y; 19 edges() {} 20 edges(int _x, int _y) : x(_x), y(_y) {} 21 }e[M]; 22 23 int n, m, k, cnt; 24 int fa[N]; 25 bool vis[M]; 26 27 char buf[Maxlen]; 28 int Len, Left; 29 30 inline int read() { 31 int x = 0; 32 while (buf[Left] < '0' || '9' < buf[Left]) ++Left; 33 while ('0' <= buf[Left] && buf[Left] <= '9') 34 x = x * 10 + buf[Left++] - '0'; 35 return x; 36 } 37 38 int len = 0, pr[15]; 39 inline void print(int x) { 40 if (x < 0) putchar('-'), x = -x; 41 while (x) 42 pr[++len] = x % 10, x /= 10; 43 if (!len) putchar('0'); 44 while (len) 45 putchar(pr[len--] + '0'); 46 } 47 48 49 int find_fa(int x) { 50 return fa[x] == x ? x : fa[x] = find_fa(fa[x]); 51 } 52 53 int main() { 54 int i, x, y, f, L, R, fa1, fa2; 55 Len = fread(buf, 1, Maxlen, stdin); 56 buf[Len] = ' '; 57 n = read(), m = read(), k = read(); 58 for (i = 1, L = 0, R = m + 1; i <= m; ++i) { 59 x = read(), y = read(), f = read(); 60 if (f) e[++L] = edges(x, y); 61 else e[--R] = edges(x, y); 62 } 63 64 for (i = 1; i <= n; ++i) fa[i] = i; 65 for (i = 1; i <= m; ++i) 66 if ((fa1 = find_fa(e[i].x)) != (fa2 = find_fa(e[i].y))) { 67 fa[fa1] = fa2; 68 if (i > L) 69 vis[i] = 1, cnt += 1; 70 } 71 if (cnt > k) { 72 puts("no solution"); 73 return 0; 74 } 75 76 for (i = 1; i <= n; ++i) fa[i] = i; 77 for (i = R; i <= m; ++i) 78 if (vis[i]) fa[find_fa(e[i].x)] = find_fa(e[i].y); 79 for (i = R; i <= m; ++i) 80 if (!vis[i] && cnt < k && (fa1 = find_fa(e[i].x)) != (fa2 = find_fa(e[i].y))) { 81 fa[fa1] = fa2, vis[i] = 1; 82 ++cnt; 83 } 84 if (cnt != k) { 85 puts("no solution"); 86 return 0; 87 } 88 89 for (i = 1; i <= L; ++i) 90 if ((fa1 = find_fa(e[i].x)) != (fa2 = find_fa(e[i].y))) 91 fa[fa1] = fa2, vis[i] = 1; 92 for (i = 1; i <= m; ++i) 93 if (vis[i]) { 94 print(e[i].x), putchar(' '); 95 print(e[i].y), putchar(' '); 96 print(i <= L), putchar(' '); 97 } 98 }
（p.s. 用IO外挂骗了个Rank.1，液！）
（p.s. 另：试了一下。。。putchar和putc差不多快，但是fputc和fwrite好慢。。。不知为何呢？）