一、Prim
题意
求最小生成树
代码
int prim()
{
memset(dist,0x3f,sizeof(dist));
int res = 0;
for(int i=0;i<n;i++){
int t= -1;
for(int j=1;j<=n;j++){
if(!st[j]&&(t==-1||dist[t]>dist[j])){
t = j;
}
}
st[t] = true;
if(i&&dist[t]==inf) return inf;
if(i) res += dist[t];
for(int j=1;j<=n;j++){
dist[j] = min(dist[j], g[t][j]);
}
}
return res;
}
二、Kruskal
题意
求最小生成树
代码
struct Edge
{
int a,b,w;
bool operator < (const Edge &W) const
{
return w < W.w;
}
}edges[M];
int find(int x)
{
if(x!=p[x]) p[x] = find(p[x]);
return p[x];
}
int kruskal()
{
sort(edges,edges+m);
for(int i=1;i<=n;i++) p[i] = i;
int res = 0, cnt = 0;
for(int i=0;i<m;i++){
auto e = edges[i];
int a = e.a, b = e.b, w = e.w;
a = find(a), b = find(b);
if(a!=b){
p[a] = b;
res += w;
cnt ++;
}
}
if(cnt<n-1) return inf;
else return res;
}
三、染色法判断二分图
题意
判断一个图是否为二分图
代码
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 100010, M = 200010, inf = 0x3f3f3f3f;
int n, m;
int h[N], e[M], ne[M], idx;
int color[N];
void add(int a,int b)
{
e[idx] = b, ne[idx] = h[a], h[a] = idx ++;
}
bool dfs(int u,int c)
{
color[u] = c;
for(int i=h[u];~i;i=ne[i]){
int j = e[i];
if(!color[j]){
if(!dfs(j,3-c)) return false;
}
else if(color[j]==color[u]) return false;
}
return true;
}
int main()
{
memset(h,-1,sizeof(h));
cin >> n >> m;
while(m--){
int a,b;
cin >> a >> b;
add(a,b), add(b,a);
}
bool flag = true;
for(int i=1;i<=n;i++){
if(!color[i]){
if(!dfs(i,1)){
flag = false;
break;
}
}
}
if(flag) puts("Yes");
else puts("No");
return 0;
}
四、匈牙利算法
题意
求二分图最大匹配
代码
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 100010, M = 200010;
int n1,n2,m;
int h[N], e[M], ne[M], idx;
int match[N];
bool st[N];
void add(int a,int b)
{
e[idx] = b, ne[idx] = h[a], h[a] = idx ++;
}
bool find(int x)
{
for(int i=h[x];~i;i=ne[i]){
int j = e[i];
if(!st[j]){
st[j] = true;
if(!match[j]||find(match[j])){
match[j] = x;
return true;
}
}
}
return false;
}
int main()
{
cin >> n1 >> n2 >> m;
memset(h,-1,sizeof(h));
while(m--){
int a,b;
cin >> a >> b;
add(a,b);
}
int res = 0;
for(int i=1;i<=n1;i++){
memset(st,false,sizeof(st));
if(find(i)) res ++;
}
cout << res << endl;
return 0;
}