今天正式开始搞图论了。
MST 裸题,prim 搞的。
#include<cstring>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN = 110;
bool vis[MAXN];
int lowc[MAXN];
int cost[MAXN][MAXN];
int Prim(int cost[][MAXN], int n){
int ans = 0;
memset(vis, false, sizeof(vis));
vis[0] = true;
for(int i = 1; i < n; i++)
lowc[i] = cost[0][i];
for(int i = 1; i < n; i++){
int minc = INF;
int p = -1;
for(int j = 0; j < n; j++){
if(!vis[j] && minc>lowc[j]){
minc = lowc[j];
p = j;
}
}
if(minc == INF) return -1;
ans += minc;
vis[p] = true;
for(int j = 0; j < n; j++){
if(!vis[j] && lowc[j]>cost[p][j])
lowc[j] = cost[p][j];
}
}
return ans;
}
int main(){
// freopen("a.txt", "r", stdin);
int n;
while(scanf("%d", &n) && n){
memset(cost, 0x3f, sizeof(cost));
int a, b, c;
for(int i = 1; i <= n*(n-1)/2; i++){
scanf("%d %d %d", &a, &b, &c);
a--; b--;
cost[a][b] = cost[b][a] = c;
}
cout << Prim(cost, n) << endl;
}
return 0;
}
Kruskal:
#include<cstring>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
const int MAXN = 110; //最大点数
const int MAXM = 10000; //最大边数
int F[MAXN]; //并查集使用
struct Edge{
int u, v, w;
}edge[MAXM]; //存储边的信息,包括起点/终点/权值
int tol; //边数,加边前赋值为0
void addedge(int u,int v,int w){
edge[tol].u = u;
edge[tol].v = v;
edge[tol++].w = w;
}
bool cmp(Edge a,Edge b){//排序函数,讲边按照权值从小到大排序
return a.w < b.w;
}
int findfa(int x){
if(F[x] == -1)return x;
else return F[x] = findfa(F[x]);
}
int Kruskal(int n){ //传入点数,返回最小生成树的权值,如果不连通返回-1
memset(F, -1, sizeof(F));
sort(edge, edge+tol, cmp);
int cnt = 0;//计算加入的边数
int ans = 0;
for(int i = 0; i < tol; i++){
int u = edge[i].u;
int v = edge[i].v;
int w = edge[i].w;
int t1 = findfa(u);
int t2 = findfa(v);
if(t1 != t2){
ans += w;
F[t1] = t2;
cnt++;
}
if(cnt == n-1)break;
}
if(cnt < n-1)return -1;//不连通
else return ans;
}
int main(){
//freopen("a.txt", "r", stdin);
int n;
while(~scanf("%d", &n) && n){
int k = n*(n-1)/2;
for(tol = 0; tol < k; ){
int a, b, c;
scanf("%d %d %d", &a, &b, &c);
addedge(a, b, c);
}
cout << Kruskal(n) << endl;
}
return 0;
}