目录
「图论」第2章 最小生成树课堂过关
A. 【例题1】繁忙都市
题目
代码
prim
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define N 310
#define M 200010
struct Heap {
#define max_(_ , __) ((_) < (__) ? (_) : (__))
int siz;
int a[M * 2] , b[M * 2];
bool ty;
inline void swap_(int x , int y) {
int tmp;
tmp = a[y] , a[y] = a[x] , a[x] = tmp;
tmp = b[y] , b[y] = b[x] , b[x] = tmp;
}
void clear() {
siz = 0 , ty = 0;
memset(a , 0 , sizeof(a));
memset(b , 0 , sizeof(b));
}
inline void push(int dat , int dat2) {
a[++siz] = dat;
b[siz] = dat2;
int p = siz;
while(p > 1 && (!(a[p >> 1] < a[p]) ^ ty))
swap_(p >> 1 , p) , p >>= 1;
}
inline void pop() {
swap_(1 , siz);
--siz;
int p = 1 , tmp;
while(p * 2 <= siz && (!(a[p] < a[tmp = ( p * 2 + 1 > siz ? p * 2 : (a[p * 2] < a[p * 2 + 1] ^ ty ? p * 2 : p * 2 + 1) ) ] ) ^ ty))
swap_(tmp , p) , p = tmp;
}
inline int top_A() {
return siz == 0 ? 0 : a[1];
}
inline int top_B() {
return siz == 0 ? 0 : b[1];
}
inline bool empty() {
return siz == 0;
}
} h;
int read() {
int re = 0;
char c = getchar();
while(c < '0' || c > '9')
c = getchar();
while(c >= '0' && c <= '9')
re = (re << 1) + (re << 3) + c - '0',
c = getchar();
return re;
}
int nxt[M] , val[M] , to[M];
int head[N];
bool vis[N];
inline void addedge(int u , int v , int value) {
static int cnt = 0;
++cnt;
val[cnt] = value , to[cnt] = v , nxt[cnt] = head[u] , head[u] = cnt;
}
int n , m;
int ans;
int main() {
n = read() , m = read();
for(int i = 1 ; i <= m ; i++) {
int u , v , value;
u = read() , v = read() , value = read();
addedge(u , v , value);
addedge(v , u , value);
}
for(int i = head[1] ; i ; i = nxt[i])
h.push(val[i] , to[i]);
vis[1] = true;
for(int j = 2 ; j <= n ; j++) {
while(vis[h.top_B()] && !h.empty())
h.pop();
if(h.empty()) break;
int u = h.top_B();
if(h.top_A() > ans)
ans = h.top_A();
h.pop();
vis[u] = true;
for(int i = head[u] ; i ; i = nxt[i]) {
if(!vis[to[i]]) {
h.push(val[i] , to[i]);
}
}
}
printf("%d %d" , n - 1 , ans);
return 0;
}
Kruskal
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
#define N 310
#define M 100010
int read() {
int re = 0;
char c = getchar();
while(c < '0' || c > '9')
c = getchar();
while(c >= '0' && c <= '9')
re = (re << 1) + (re << 3) + c - '0',
c = getchar();
return re;
}
struct ednode{
int u , v , val;
}ed[M];
bool cmp(ednode a , ednode b) {
return a.val < b.val;
}
int fa[N];
int findroot(int x) {
return fa[x] == x ? x : (fa[x] = findroot(fa[x]));
}
void uni(int x , int y) {
fa[findroot(x)] = findroot(y);
}
int n , m;
int ans;
int main() {
n = read() , m = read();
for(int i = 1 ; i <= m ; i++)
ed[i].u = read() , ed[i].v = read() , ed[i].val = read();
sort(ed + 1 , ed + m + 1 , cmp);
for(int i = 1 ; i <= n ; i++)
fa[i] = i;
for(int i = 1 ; i <= m ; i++) {
if(findroot(ed[i].u) != findroot(ed[i].v)) {
uni(ed[i].u , ed[i].v);
if(ed[i].val > ans)
ans = ed[i].val;
}
}
printf("%d %d" , n - 1 , ans);
return 0;
}
B. 【例题2】新的开始
题目
思路
想象一个原点,它到(i)点的距离就是(i)建发电站的费用,其余连边不变,从那个原点开始跑一边最小生成树即可
代码
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define N 310
#define M N * N
struct Heap {
#define max_(_ , __) ((_) < (__) ? (_) : (__))
int siz;
int a[M * 2] , b[M * 2];
bool ty;
inline void swap_(int x , int y) {
int tmp;
tmp = a[y] , a[y] = a[x] , a[x] = tmp;
tmp = b[y] , b[y] = b[x] , b[x] = tmp;
}
void clear() {
siz = 0 , ty = 0;
memset(a , 0 , sizeof(a));
memset(b , 0 , sizeof(b));
}
inline void push(int dat , int dat2) {
a[++siz] = dat;
b[siz] = dat2;
int p = siz;
while(p > 1 && (!(a[p >> 1] < a[p]) ^ ty))
swap_(p >> 1 , p) , p >>= 1;
}
inline void pop() {
swap_(1 , siz);
--siz;
int p = 1 , tmp;
while(p * 2 <= siz && (!(a[p] < a[tmp = ( p * 2 + 1 > siz ? p * 2 : (a[p * 2] < a[p * 2 + 1] ^ ty ? p * 2 : p * 2 + 1) ) ] ) ^ ty))
swap_(tmp , p) , p = tmp;
}
inline int top_A() {
return siz == 0 ? 0 : a[1];
}
inline int top_B() {
return siz == 0 ? 0 : b[1];
}
inline bool empty() {
return siz == 0;
}
} h;
int read() {
int re = 0;
char c = getchar();
while(c < '0' || c > '9')
c = getchar();
while(c >= '0' && c <= '9')
re = (re << 1) + (re << 3) + c - '0',
c = getchar();
return re;
}
int n , m;
int ans;
int map[N][N];
bool vis[N];
int main() {
n = read();
for(int i = 1 ; i <= n ; i++)
h.push(read() , i);
for(int i = 1 ; i <= n ; i++)
for(int j = 1 ; j <= n ; j++) {
map[i][j] = read();
}
for(int j = 1 ; j <= n ; j++) {
while(vis[h.top_B()] && !h.empty())
h.pop();
if(h.empty()) break;
int u = h.top_B();
ans += h.top_A();
h.pop();
vis[u] = true;
for(int i = 1 ; i <= n ; i++) {
if(!vis[i])
h.push(map[i][u] , i);
}
}
printf("%d" , ans);
return 0;
}
C. 【例题3】公路建设
题目
思路
题目的意思是给定(m)条边,依次加入一开始没有边的图(G)中,若当前加入的边不能使(G)强连通,输出"0",否则,输出图(G)的最小生成树
我们把边依次加入一条链中,按插入排序的思想维护链的有序性,跑一次Kruskal就可以求出当前的最小生成树
时间复杂度(O(ncdot m))
代码
#include <iostream>
#include <cstdio>
#define N 510
#define M 2010
using namespace std;
int read() {
int re = 0;
char c = getchar();
while(c < '0' || c > '9')
c = getchar();
while(c >= '0' && c <= '9')
re = (re << 1) + (re << 3) + c - '0',
c = getchar();
return re;
}
struct ednode {
int x , y , val , nxt;
}ed[M];
int head = 1;
void insert(int x , int y , int val) {
static int cnt = 0;
++cnt;
ed[cnt].x = x , ed[cnt].y = y , ed[cnt].val = val;
if(cnt == 1) return;
int p = head;
if(val < ed[p].val) {
ed[cnt].nxt = head , head = cnt;
return;
}
for( ; ed[p].nxt != 0 ; p = ed[p].nxt)
if(ed[p].val <= val && val <= ed[ed[p].nxt].val) {
ed[cnt].nxt = ed[p].nxt , ed[p].nxt = cnt;
return;
}
ed[p].nxt = cnt;
ed[cnt].nxt = 0;
return;
}
int n , m;
int fa[N];
int findroot(int x) {
return fa[x] == x ? x : (fa[x] = findroot(fa[x]));
}
void uni(int x , int y) {
fa[findroot(x)] = findroot(y);
}
int main() {
n = read() , m = read();
for(int i = 1 ; i <= m ; i++) {
int u = read() , v = read() , val = read();
insert(u , v , val);
for(int j = 1 ; j <= n ; j++)
fa[j] = j;
int ans = 0;
for(int j = head ; j ; j = ed[j].nxt)
if(findroot(ed[j].x) != findroot(ed[j].y)) {
uni(ed[j].x , ed[j].y);
ans += ed[j].val;
}
int ty = true;
int sta = findroot(1);
for(int i = 1 ; i <= n ; i++)
if(findroot(i) != sta) {
ty = false;
break;
}
if(!ty)
puts("0");
else
printf("%.1f
" , ans * 0.5);
}
return 0;
}
D. 【例题4】构造完全图
题目
思路
50分
有一个贪心:如果(u)到(v)路径上最大边权为(val),则(u),(v)之间连一条边权小于等于(val)的边,原最小生成树会改变,否则不会改变.具体的证明可以从边权及边的数量出发,这里不再赘述
因此,直接枚举每个点,各跑一边深搜遍历整棵树即可,时间复杂度(O(n^2))
100分
从数据规模可以看到,(n)去到(10^5),即边的数量可以达到(10^{10}),一条一条算肯定是不行的.
考虑对于最小生成树的每一条边(ed_i),它连接两个连通块(T_1),(T_2),大小为(val),它一定是连接(T_1),(T_2)所有边中最小的,设(T_1)中有(siz_1)个节点,(T_2)中有(siz_2)个节点,则除了(ed_i),一共有(siz_1 cdot siz_2-1)条,每条(除(ed_i)外)的最小权值为(val+1)
更具体地,代码解释吧
代码
50分
#include <iostream>
#include <cstdio>
#define ll long long
#define N 100010
using namespace std;
int read() {
int re = 0;
char c = getchar();
while(c < '0' || c > '9')
c = getchar();
while(c >= '0' && c <= '9')
re = (re << 1) + (re << 3) + c - '0',
c = getchar();
return re;
}
struct ednode {
int to , val , nxt;
}ed[N * 2];
int head[N];
inline void addedge(int x , int y , int v) {
static int cnt = 0;
++cnt;
ed[cnt].to = y , ed[cnt].val = v , ed[cnt].nxt = head[x] , head[x] = cnt;
return;
}
int n;
ll ans;
void dfs(int x , int maxv , int fa , int dep) {
if(dep > 1) ans += maxv + 1;
for(int i = head[x] ; i ; i = ed[i].nxt) {
if(ed[i].to == fa) continue;
dfs(ed[i].to , maxv > ed[i].val ? maxv : ed[i].val , x , dep + 1);
}
}
int main() {
n = read();
for(int i = 1 ; i < n ; i++) {
int u = read() , v = read() , val;
ans += (val = read()) * 2;
addedge(u , v , val);
addedge(v , u , val);
}
for(int i = 1 ; i <= n ; i++) {
dfs(i , -1 , i , 0);
}
cout << ans / 2;
return 0;
}
100分
#include <iostream>
#include <cstdio>
#include <algorithm>
#define N 100010
using namespace std;
int read() {
int re = 0;
char c = getchar();
while(c < '0' || c > '9')
c = getchar();
while(c >= '0' && c <= '9')
re = (re << 1) + (re << 3) + c - '0',
c = getchar();
return re;
}
struct ednode{
int u , v , val;
}ed[N * 2];
bool cmp(ednode a , ednode b) {
return a.val < b.val;
}
int fa[N] , siz[N];
int findroot(int x) {
return x == fa[x] ? x : (fa[x] = findroot(fa[x]));
}
int n;
long long ans;
int main() {
n = read();
for(int i = 1 ; i < n ; i++)
ed[i].u = read() , ed[i].v = read() , ans += (ed[i].val = read());
sort(ed + 1 , ed + n , cmp);
for(int i = 1 ; i <= n ; i++)
fa[i] = i , siz[i] = 1;
for(int i = 1 ; i < n ; i++) {
int u = findroot(ed[i].u) , v = findroot(ed[i].v);
if(u != v) {
ans += (long long)(siz[u] * siz[v] - 1) * (ed[i].val + 1);
fa[u] = v;
siz[v] += siz[u];
}
}
cout << ans;
return 0;
}
随机数据生成
#include <bits/stdc++.h>
using namespace std;
int random(int r , int l = 1) {
return (long long) rand() * rand() % (r - l + 1) + l;
}
pair <int,int> ed[1000010];
int dict[1000010];//用于打乱结点编号,有些有向图不适用
int main() {
unsigned seed;
cin >> seed;
seed *= time(0);
srand(seed);//外界输入种子
int n = random(1e4 , 5);
for(int i = 1 ; i <= n ; i++)
dict[i] = i;
random_shuffle(dict + 1 , dict + n + 1);
for(int i = 1 ; i < n ; i++) {
ed[i].first = random(i);
ed[i].second = i + 1;
}
printf("%d
" , n);
random_shuffle(ed + 1 , ed + n);//随机打乱
for(int i = 1 ;i < n ; i++)
printf("%d %d %d
" , dict[ed[i].first] , dict[ed[i].second] , random(1e5));
return 0;
}