题目链接：https://www.lydsy.com/JudgeOnline/problem.php?id=1196
看起来和免费道路那道题是有共同点的，但细细一想，其实二者并不一样。免费道路要求恰好有k条鹅卵石路，而本题要求至少k条一级道路。
本题是想让边权最大的边最小（别去想最小生成树的性质！！！），所以可以二分答案，因为必然最大边权越大越容易满足，任意一条边都可以成为一级道路。
我们毕竟是要验证答案，所以应该先加一遍一级道路，然后再去补二级道路，这样才能够保证符合条件的不被判断为不符合条件。
1 #include <cstdio> 2 #include <algorithm> 3 4 using namespace std; 5 6 inline int get_num() { 7 int num = 0; 8 char c = getchar(); 9 while (c < '0' || c > '9') c = getchar(); 10 while (c >= '0' && c <= '9') 11 num = num * 10 + c - '0', c = getchar(); 12 return num; 13 } 14 15 const int maxn = 1e4 + 5, maxm = 2e4 + 5; 16 17 struct Edge { 18 int u, v, w1, w2; 19 } edge[maxm]; 20 21 int n, k, m, maxw, fa[maxn]; 22 23 int dj_find(int i) { 24 if (fa[i] == i) return i; 25 else return fa[i] = dj_find(fa[i]); 26 } 27 28 inline void dj_merge(int a, int b) { 29 fa[dj_find(a)] = dj_find(b); 30 } 31 32 inline int check(int x) { 33 int cnt = 0; 34 for (int i = 1; i <= n; ++i) fa[i] = i; 35 for (int i = 1; i <= m - 1; ++i) { 36 if (edge[i].w1 > x) continue; 37 int u = edge[i].u, v = edge[i].v; 38 if (dj_find(u) != dj_find(v)) { 39 dj_merge(u, v); 40 ++cnt; 41 } 42 } 43 if (cnt < k) return 0; 44 for (int i = 1; i <= m - 1; ++i) { 45 if (edge[i].w2 > x) continue; 46 int u = edge[i].u, v = edge[i].v; 47 if (dj_find(u) != dj_find(v)) { 48 dj_merge(u, v); 49 ++cnt; 50 } 51 } 52 if (cnt != n - 1) return 0; 53 else return 1; 54 } 55 56 int main() { 57 n = get_num(), k = get_num(), m = get_num(); 58 for (int i = 1; i <= m - 1; ++i) { 59 edge[i].u = get_num(), edge[i].v = get_num(); 60 edge[i].w1 = get_num(), edge[i].w2 = get_num(); 61 maxw = max(maxw, edge[i].w1); 62 } 63 int l = 1, r = maxw; 64 while (l < r) { 65 int mid = l + (r - l) / 2; 66 if (check(mid)) r = mid; 67 else l = mid + 1; 68 } 69 printf("%d", l); 70 return 0; 71 }
然而，做完一道题，刷新世界观。。。
本着骗一下AC量的原则，我把相同的代码交到了洛谷上，，，结果，呵呵呵。。。
一看，哦输出格式变了。我就稍微一改，死活过不去，用Special Judge评，老是说我图没连通。那我快去看看题解吧。
居然，还可以用贪心做！那我改去学学贪心好了。。。
参照Kruskal算法的做法，我们先按c1从小到大排一下序，然后从里面条k条，肯定刚好k条最好；然后选择二级道路肯定会更好些，所以我们再按照c2排序，挑完剩下的。
注意，并不保证最小的c1大于最大的c2；若c1相同则按c2从大到小排序，因为这样可以保证后面最优。
1 #include <cstdio> 2 #include <algorithm> 3 4 using namespace std; 5 6 inline int get_num() { 7 int num = 0; 8 char c = getchar(); 9 while (c < '0' || c > '9') c = getchar(); 10 while (c >= '0' && c <= '9') 11 num = num * 10 + c - '0', c = getchar(); 12 return num; 13 } 14 15 const int maxn = 1e4 + 5, maxm = 2e4 + 5; 16 17 struct Edge { 18 int id, u, v, w1, w2; 19 } edge[maxm]; 20 21 bool comp1(const Edge& lhs, const Edge& rhs) { 22 if (lhs.w1 == rhs.w1) return lhs.w2 > rhs.w2; 23 else return lhs.w1 < rhs.w1; 24 } //w1相等，则留w2小的，之后会更优 25 26 bool comp2(const Edge& lhs, const Edge& rhs) { 27 return lhs.w2 < rhs.w2; 28 } 29 30 int fa[maxn]; 31 32 int dj_find(int i) { 33 if (fa[i] == i) return i; 34 else return fa[i] = dj_find(fa[i]); 35 } 36 37 inline void dj_merge(int a, int b) { 38 fa[dj_find(a)] = dj_find(b); 39 } 40 41 struct Ans { 42 int id, type; 43 bool operator < (const Ans& rhs) const { 44 return id < rhs.id; 45 } 46 } ans[maxn]; 47 48 int main() { 49 int n = get_num(), k = get_num(), m = get_num(); 50 int cnt = 0, me = 0, aeid = 0; 51 for (int i = 1; i <= m - 1; ++i) { 52 edge[i].id = i; 53 edge[i].u = get_num(), edge[i].v = get_num(); 54 edge[i].w1 = get_num(), edge[i].w2 = get_num(); 55 } 56 sort(edge + 1, edge + m, comp1); 57 for (int i = 1; i <= n; ++i) fa[i] = i; 58 for (int i = 1; i <= m - 1; ++i) { 59 int u = edge[i].u, v = edge[i].v; 60 if (dj_find(u) != dj_find(v)) { 61 dj_merge(u, v); 62 me = max(me, edge[i].w1); 63 ans[++aeid].id = edge[i].id, ans[aeid].type = 1; 64 if (++cnt == k) break; 65 } 66 } 67 sort(edge + 1, edge + m, comp2); 68 for (int i = 1; i <= m - 1; ++i) { 69 int u = edge[i].u, v = edge[i].v; 70 if (dj_find(u) != dj_find(v)) { 71 dj_merge(u, v); 72 me = max(me, edge[i].w2); 73 ans[++aeid].id = edge[i].id, ans[aeid].type = 2; 74 } 75 } 76 printf("%d ", me); 77 sort(ans + 1, ans + n); 78 for (int i = 1; i <= n - 1; ++i) 79 printf("%d %d ", ans[i].id, ans[i].type); 80 return 0; 81 }
当然，并不是说二分不能做，二分也可以，是我写错了而已。
我以为跑完二分，顺便记录每一次的答案即可，但显然不对，因为你不知道最后一次是对的还是错的，所以我们需要在求出答案之后，再建一次树。
1 #include <cstdio> 2 #include <algorithm> 3 4 using namespace std; 5 6 inline int get_num() { 7 int num = 0; 8 char c = getchar(); 9 while (c < '0' || c > '9') c = getchar(); 10 while (c >= '0' && c <= '9') 11 num = num * 10 + c - '0', c = getchar(); 12 return num; 13 } 14 15 const int maxn = 1e4 + 5, maxm = 2e4 + 5; 16 17 struct Edge { 18 int u, v, w1, w2; 19 } edge[maxm]; 20 21 int n, k, m, maxw, fa[maxn]; 22 23 int dj_find(int i) { 24 if (fa[i] == i) return i; 25 else return fa[i] = dj_find(fa[i]); 26 } 27 28 inline void dj_merge(int a, int b) { 29 fa[dj_find(a)] = dj_find(b); 30 } 31 32 struct Ans { 33 int id, type; 34 bool operator < (const Ans& rhs) const { 35 return id < rhs.id; 36 } 37 } ans[maxn]; 38 39 inline int check(int x) { 40 int cnt = 0; 41 for (int i = 1; i <= n; ++i) fa[i] = i; 42 for (int i = 1; i <= m - 1; ++i) { 43 if (edge[i].w1 > x) continue; 44 int u = edge[i].u, v = edge[i].v; 45 if (dj_find(u) != dj_find(v)) { 46 dj_merge(u, v); 47 ++cnt; 48 } 49 } 50 if (cnt < k) return 0; 51 for (int i = 1; i <= m - 1; ++i) { 52 if (edge[i].w2 > x) continue; 53 int u = edge[i].u, v = edge[i].v; 54 if (dj_find(u) != dj_find(v)) { 55 dj_merge(u, v); 56 ++cnt; 57 } 58 } 59 if (cnt != n - 1) return 0; 60 else return 1; 61 } 62 63 inline void build(int x) { 64 int cnt = 0; 65 for (int i = 1; i <= n; ++i) fa[i] = i; 66 for (int i = 1; i <= m - 1; ++i) { 67 if (edge[i].w1 > x) continue; 68 int u = edge[i].u, v = edge[i].v; 69 if (dj_find(u) != dj_find(v)) { 70 dj_merge(u, v); 71 ans[++cnt].id = i, ans[cnt].type = 1; 72 } 73 } 74 for (int i = 1; i <= m - 1; ++i) { 75 if (edge[i].w2 > x) continue; 76 int u = edge[i].u, v = edge[i].v; 77 if (dj_find(u) != dj_find(v)) { 78 dj_merge(u, v); 79 ans[++cnt].id = i, ans[cnt].type = 2; 80 } 81 } 82 } 83 84 int main() { 85 n = get_num(), k = get_num(), m = get_num(); 86 for (int i = 1; i <= m - 1; ++i) { 87 edge[i].u = get_num(), edge[i].v = get_num(); 88 edge[i].w1 = get_num(), edge[i].w2 = get_num(); 89 maxw = max(maxw, edge[i].w1); 90 } 91 int l = 1, r = maxw; 92 while (l < r) { 93 int mid = l + (r - l) / 2; 94 if (check(mid)) r = mid; 95 else l = mid + 1; 96 } 97 printf("%d ", l); 98 build(l); 99 sort(ans + 1, ans + n); 100 for (int i = 1; i <= n - 1; ++i) 101 printf("%d %d ", ans[i].id, ans[i].type); 102 return 0; 103 }