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。

#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 + ;
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 = ; 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 = ;
nowAns = ;
struct Edge t;
t.w = val;
int pos = lower_bound(e + , e + + 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 - ) {
return true;
}
}
return false;
}
void work() {
scanf("%d%d", &n, &m);
LL lo = 1e18L, hi = -1e18L;
for (int i = ; 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 + , e + + m);
while (lo <= hi) {
LL mid = (lo + hi) >> ;
if (check(mid)) {
hi = mid - ;
} else lo = mid + ;
}
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 ;
}

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。

#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 + ;
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 = ; i <= n; ++i) fa[i] = i;
}
int result() {
init();
int sel = ;
for (int i = ; i <= m; ++i) {
if (tofind(e[i].u) == tofind(e[i].v)) continue;
tomerge(e[i].u, e[i].v);
sel++;
if (sel == n - ) return e[i].w;
}
}
void work() {
scanf("%d%d", &n, &m);
for (int i = ; i <= m; ++i) {
scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w);
}
sort(e + , e + + m);
int mx = result();
sort(e + , e + + m, cmp);
init();
LL ans = ;
int sel = ;
for (int i = ; 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 - ) 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 ;
}
05-11 22:39