51nod 1640 天气晴朗的魔法 二分 + 克鲁斯卡算法(kruskal算法) 做复杂了
http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1640
一开始想的时候,看到要使得最大值最小,那这样肯定是二分这个最大值了,然后每一次都跑一次kruskal
这样的复杂度是O(E * 64),然后被卡TLE了
然后观察到kruskal的时候,如果最大边是val,那么比val大的是不要的了,然后整个数组也是有序的。
比如7、6、5、4、3、2、1等,这个也是可以lower_bound的,然后lower_bound后就能过,600ms
改了lower_bound后,我忘记删除那个if了,居然还是超时。TAT,这数据有点强。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <algorithm> #include <assert.h> #define IOS ios::sync_with_stdio(false) using namespace std; #define inf (0x3f3f3f3f) typedef long long int LL; #include <iostream> #include <sstream> #include <vector> #include <set> #include <map> #include <queue> #include <string> #include <bitset> int n, m; const int maxn = 2e5 + 20; struct Edge { int u, v, w; bool operator < (const struct Edge & rhs) const { return w > rhs.w; } }e[maxn]; int fa[maxn]; void init() { for (int i = 1; i <= n; ++i) fa[i] = i; } int tofind(int u) { if (fa[u] == u) return u; else return fa[u] = tofind(fa[u]); } void tomerge(int x, int y) { x = tofind(x); y = tofind(y); fa[y] = x; } LL nowAns; bool check(LL val) { init(); int sel = 0; nowAns = 0; struct Edge t; t.w = val; int pos = lower_bound(e + 1, e + 1 + m, t) - e; for (int i = pos; i <= m; ++i) { // if (e[i].w > val) continue; 卡TLE if (tofind(e[i].u) == tofind(e[i].v)) continue; tomerge(e[i].u, e[i].v); nowAns += e[i].w; sel++; if (sel == n - 1) { return true; } } return false; } void work() { scanf("%d%d", &n, &m); LL lo = 1e18L, hi = -1e18L; for (int i = 1; i <= m; ++i) { scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w); lo = min(lo, (LL)e[i].w); hi = max(hi, (LL)e[i].w); } sort(e + 1, e + 1 + m); while (lo <= hi) { LL mid = (lo + hi) >> 1; if (check(mid)) { hi = mid - 1; } else lo = mid + 1; } check(lo); // printf("%lld\n", nowAns); cout << nowAns << endl; // struct Edge t; // t.w = 3; // int pos = lower_bound(e + 1, e + 1 + m, t) - e; // cout << pos << endl; } int main() { #ifdef local freopen("data.txt", "r", stdin); // freopen("data.txt", "w", stdout); #endif work(); return 0; }
然后就上网看题解了,然后发现自己做法很复杂了。
感觉是题意弄得我们想复杂了,又魔法连的值,又总和,
正解是kruskal两次,第一次就能找出最大值最小,然后从大的再kruskal一次。
但是为什么还是700ms,有点坑。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <algorithm> #include <assert.h> #define IOS ios::sync_with_stdio(false) using namespace std; #define inf (0x3f3f3f3f) typedef long long int LL; #include <iostream> #include <sstream> #include <vector> #include <set> #include <map> #include <queue> #include <string> #include <bitset> int n, m; const int maxn = 2e5 + 20; struct Node { int u, v, w; bool operator < (const struct Node & rhs) const { return w < rhs.w; } }e[maxn]; bool cmp(struct Node a, struct Node b) { return a.w > b.w; } int fa[maxn]; int tofind(int u) { if (fa[u] == u) return u; else return fa[u] = tofind(fa[u]); } void tomerge(int x, int y) { x = tofind(x); y = tofind(y); fa[y] = x; } void init() { for (int i = 1; i <= n; ++i) fa[i] = i; } int result() { init(); int sel = 0; for (int i = 1; i <= m; ++i) { if (tofind(e[i].u) == tofind(e[i].v)) continue; tomerge(e[i].u, e[i].v); sel++; if (sel == n - 1) return e[i].w; } } void work() { scanf("%d%d", &n, &m); for (int i = 1; i <= m; ++i) { scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w); } sort(e + 1, e + 1 + m); int mx = result(); sort(e + 1, e + 1 + m, cmp); init(); LL ans = 0; int sel = 0; for (int i = 1; i <= m; ++i) { if (e[i].w > mx) continue; if (tofind(e[i].u) == tofind(e[i].v)) continue; tomerge(e[i].u, e[i].v); sel++; ans += e[i].w; if (sel == n - 1) break; } printf("%lld\n", ans); // cout << ans << endl; } int main() { #ifdef local freopen("data.txt", "r", stdin); // freopen("data.txt", "w", stdout); #endif work(); return 0; }
既然选择了远方,就要风雨兼程~
posted on 2017-03-20 18:14 stupid_one 阅读(165) 评论(0) 编辑 收藏 举报