如题，给出一个无向图，求出最小生成树，如果该图不连通，则输出 orz。
输入格式
第一行包含两个整数 N,M，表示该图共有 N 个结点和 M 条无向边。
接下来 M行每行包含三个整数 X_i,Y_i,Z_i，表示有一条长度为 Z_i的无向边连接结点 X_i,Y_i
输出格式
如果该图连通，则输出一个整数表示最小生成树的各边的长度之和。如果该图不连通则输出 orz。
输入输出样例
输入
4 5
1 2 2
1 3 2
1 4 3
2 3 4
3 4 3
输出
7
数据规模：
对于 100% 的数据：\(1\le N\le 5000 ,1\le M\le 2\times 10^5\)
prim
#include <bits/stdc++.h>
namespace FastIO {
char buf[1 << 21], buf2[1 << 21], a[20], *p1 = buf, *p2 = buf, hh = '\n';
int p, p3 = -1;
void read() {}
void print() {}
inline int getc() {
return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
}
inline void flush() {
fwrite(buf2, 1, p3 + 1, stdout), p3 = -1;
}
template<typename T, typename... T2>
inline void read(T &x, T2 &... oth) {
int f = 0;
x = 0;
char ch = getc();
while (!isdigit(ch)) {
if (ch == '-')
f = 1;
ch = getc();
}
while (isdigit(ch)) {
x = x * 10 + ch - 48;
ch = getc();
}
x = f ? -x : x;
read(oth...);
}
template<typename T, typename... T2>
inline void print(T x, T2... oth) {
if (p3 > 1 << 20)
flush();
if (x < 0)
buf2[++p3] = 45, x = -x;
do {
a[++p] = x % 10 + 48;
} while (x /= 10);
do {
buf2[++p3] = a[p];
} while (--p);
buf2[++p3] = hh;
print(oth...);
}
} // namespace FastIO
#define read FastIO::read
#define print FastIO::print
//======================================
using namespace std;
const int maxn=5e3+10;
const int inf=0x3f3f3f3f;
typedef long long ll;
int n,m,e[maxn][maxn],ans,vis[maxn],d[maxn];
int main() {
#ifndef ONLINE_JUDGE
freopen("1.txt", "r", stdin);
//freopen("2.txt", "w", stdout);
#endif
//======================================
read(n, m);
memset(e,0x3f,sizeof(e));
for (int i = 1, u, v, w; i <= m; i++) {
read(u, v, w);
e[u][v] = min(e[u][v], w);
e[v][u] = min(e[v][u], w);
}
memset(d, 0x3f, sizeof(d));
for (int i = 1; i <= n; i++) {
d[i] = e[1][i];
}
vis[1]=1;
for (int i = 1; i <= n; i++) {
int mi = inf, miv = -1;
for (int j = 1; j <= n; j++) {
if (!vis[j] && d[j] < mi) {
mi = d[j];
miv = j;
}
}
if (miv == -1) break;
vis[miv] = 1;
ans += mi;
//print(mi);
for (int j = 1; j <= n; j++) {
if (!vis[j]) {
d[j] = min(d[j], e[miv][j]);
}
}
}
print(ans);
//======================================
FastIO::flush();
return 0;
}
堆优化的prim
#include <bits/stdc++.h>
namespace FastIO {
char buf[1 << 21], buf2[1 << 21], a[20], *p1 = buf, *p2 = buf, hh = '\n';
int p, p3 = -1;
void read() {}
void print() {}
inline int getc() {
return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
}
inline void flush() {
fwrite(buf2, 1, p3 + 1, stdout), p3 = -1;
}
template<typename T, typename... T2>
inline void read(T &x, T2 &... oth) {
int f = 0;
x = 0;
char ch = getc();
while (!isdigit(ch)) {
if (ch == '-')
f = 1;
ch = getc();
}
while (isdigit(ch)) {
x = x * 10 + ch - 48;
ch = getc();
}
x = f ? -x : x;
read(oth...);
}
template<typename T, typename... T2>
inline void print(T x, T2... oth) {
if (p3 > 1 << 20)
flush();
if (x < 0)
buf2[++p3] = 45, x = -x;
do {
a[++p] = x % 10 + 48;
} while (x /= 10);
do {
buf2[++p3] = a[p];
} while (--p);
buf2[++p3] = hh;
print(oth...);
}
} // namespace FastIO
#define read FastIO::read
#define print FastIO::print
//======================================
using namespace std;
const int maxn=4e5+10;
const int inf=0x3f3f3f3f;
typedef long long ll;
int n,m,ans,vis[maxn],head[maxn];
priority_queue<pair<int,int> ,vector<pair<int,int> >,greater<pair<int,int> > >q;
struct node{
int v,nxt,w;
}e[maxn];
int t;
void add(int u,int v,int w) {
t++;
e[t].v = v;
e[t].w = w;
e[t].nxt = head[u];
head[u] = t;
}
int main() {
#ifndef ONLINE_JUDGE
freopen("1.txt", "r", stdin);
//freopen("2.txt", "w", stdout);
#endif
//======================================
read(n, m);
for (int i = 1, u, v, w; i <= m; i++) {
read(u, v, w);
add(u, v, w);
add(v, u, w);
}
int cnt=0;
q.push(make_pair(0, 1));
while (!q.empty()) {
int w = q.top().first, u = q.top().second;
q.pop();
if (vis[u]) continue;
vis[u] = 1;
ans += w;
cnt++;
if (cnt>=n) break;
for (int i = head[u]; i; i = e[i].nxt) {
int v=e[i].v;
if (!vis[v]) q.push(make_pair(e[i].w, v));
}
}
print(ans);
//======================================
FastIO::flush();
return 0;
}
Kruskal
#include <bits/stdc++.h>
namespace FastIO {
char buf[1 << 21], buf2[1 << 21], a[20], *p1 = buf, *p2 = buf, hh = '\n';
int p, p3 = -1;
void read() {}
void print() {}
inline int getc() {
return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
}
inline void flush() {
fwrite(buf2, 1, p3 + 1, stdout), p3 = -1;
}
template<typename T, typename... T2>
inline void read(T &x, T2 &... oth) {
int f = 0;
x = 0;
char ch = getc();
while (!isdigit(ch)) {
if (ch == '-')
f = 1;
ch = getc();
}
while (isdigit(ch)) {
x = x * 10 + ch - 48;
ch = getc();
}
x = f ? -x : x;
read(oth...);
}
template<typename T, typename... T2>
inline void print(T x, T2... oth) {
if (p3 > 1 << 20)
flush();
if (x < 0)
buf2[++p3] = 45, x = -x;
do {
a[++p] = x % 10 + 48;
} while (x /= 10);
do {
buf2[++p3] = a[p];
} while (--p);
buf2[++p3] = hh;
print(oth...);
}
} // namespace FastIO
#define read FastIO::read
#define print FastIO::print
//======================================
using namespace std;
const int maxn=2e5+10;
const int inf=0x3f3f3f3f;
typedef long long ll;
int n,m,ans,f[maxn];
struct node{
int u,v,w;
}e[maxn];
int find(int x) {
return x == f[x] ? x : f[x] = find(f[x]);
}
bool cmp(const node &a,const node &b) {
return a.w < b.w;
}
int main() {
#ifndef ONLINE_JUDGE
freopen("1.txt", "r", stdin);
//freopen("2.txt", "w", stdout);
#endif
//======================================
read(n, m);
for (int i = 1; i <= m; i++) {
read(e[i].u, e[i].v, e[i].w);
}
for (int i = 1; i <= n; i++) {
f[i] = i;
}
sort(e + 1, e + m + 1, cmp);
for (int i = 1; i <= m; i++) {
int u = find(e[i].u), v = find(e[i].v);
if (u == v) continue;
ans += e[i].w;
f[u] = v;
}
print(ans);
//======================================
FastIO::flush();
return 0;
}