【ACM】- HDU.1863 暢通工程 【最小生成樹】
阿新 • • 發佈:2018-12-09
題目連結
題目分析
最小生成樹問題,求路徑和
解題思路
算是最小生成樹的母題,分別用以下幾種方法實現以下:
1、Kruskal
演算法 + 並查集;
2、Prime
演算法 (鄰接矩陣版本)
3、Prime
演算法(鄰接表版本)
分別再用堆結構(priority_queue
)優化一下
| Kruskal
演算法 + 並查集 (堆優化priority_queue
)
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn = 110;
struct edge{
int u, v, cost;
edge() {}
edge(int _u, int _v, int _cost) : u(_u), v(_v), cost(_cost) {} //建構函式,便於加入結點
bool operator < (const edge& n) const { //規定優先順序
return cost > n.cost; //注意和sort函式是相反的
}
};
int N, M;
int far[maxn]; //並查集
//尋根
int find_root(int a) {
int root = a;
while(root != far[root]) root = far[root];
while(a != far[a]) { //路徑壓縮
int cur = a;
a = far[a];
far[cur] = root;
}
return root;
}
//合併集合
void union_set(int a, int b) {
int root_a = find_root(a);
int root_b = find_root(b);
if (a != b){
far[root_b] = root_a;
}
}
int kruskal(priority_queue<edge> E) {
for(int i = 1; i <= N; i++) far[i] = i; //初始化並查集
int ans = 0;//權值和
int edge_num = 0; //已選擇的邊數
int cnt = N; //連通塊數
for(int i = 0; i < M; i++) {
edge e = E.top(); E.pop(); //get fisrt edge
int root_u = find_root(e.u);
int root_v = find_root(e.v);
if(root_u != root_v) {
union_set(root_u, root_v);
edge_num++;
cnt--; //連通塊數-1
ans += e.cost;
}
if(edge_num == N - 1) break; //邊數等於結點數-1
}
if(cnt != 1) return -1;//只剩一個連通塊(edge_num == N - 1 也沒問題)
else return ans;
}//kruskal
int main() {
int a, b, cost;
while(scanf("%d %d", &M, &N) != EOF) {
if(M == 0) break;
priority_queue<edge> E; //儲存所有邊(無clear()函式,每次重新定義時間最快)
//優先順序和sort()函式是相反的
for(int i = 0; i < M; i++) {
scanf("%d %d %d", &a, &b, &cost);
E.push(edge(a, b, cost)); //加入堆
}
int ans = kruskal(E);
if(ans == -1) printf("?\n");
else printf("%d\n", ans);
}//while
system("pause");
return 0;
}
| Kruskal
演算法 + 並查集 (sort())
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn = 110;
int N, M;
int far[maxn]; //並查集
struct edge{
int u, v, cost;
bool operator < (const edge& n) const { //規定優先順序
return cost < n.cost;
}
}E[maxn];
bool cmp(edge e, edge f) { //也可以用自定義比較函式
return e.cost < f.cost;
}
//尋根
int find_root(int a) {
int root = a;
while(root != far[root]) root = far[root];
while(a != far[a]) { //路徑壓縮
int cur = a;
a = far[a];
far[cur] = root;
}
return root;
}
//合併集合
void union_set(int a, int b) {
int root_a = find_root(a);
int root_b = find_root(b);
if(a != b){
far[root_b] = root_a;
}
}
int kruskal() {
for(int i = 1; i <= N; i++) far[i] = i; //初始化並查集
sort(E, E + M); //邊遞增排序(也可直接用堆實現priority_queue)
//sort(E, E + M, cmp);
int ans = 0;//權值和
int edge_num = 0; //已選擇的邊數
int cnt = N; //連通塊數
for(int i = 0; i < M; i++) {
int root_u = find_root(E[i].u);
int root_v = find_root(E[i].v);
if(root_u != root_v) {
union_set(root_u, root_v);
edge_num++;
cnt--; //連通塊數-1
ans += E[i].cost;
}
if(edge_num == N - 1) break; //邊數等於結點數-1
}
if(cnt != 1) return -1;//只剩一個連通塊
else return ans;
}//kruskal
int main() {
int a, b, cost;
while(scanf("%d %d", &M, &N) != EOF) {
if(M == 0) break;
for(int i = 0; i < M; i++) {
scanf("%d %d %d", &a, &b, &cost);
E[i].u = a; E[i].v = b;
E[i].cost = cost;
}
int ans = kruskal();
if(ans == -1) printf("?\n");
else printf("%d\n", ans);
}//while
system("pause");
return 0;
}
| Prime
演算法(鄰接表版本)
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
const int INF = 0x3fffffff;
const int maxn = 110;
int N, M;
int G[maxn][maxn];
int d[maxn]; //Prime
bool vis[maxn];
//Prime演算法
int prime(int st) {
fill(d, d + maxn, INF);
fill(vis, vis + maxn, false);
int ans = 0;
d[st] = 0; //起點(根)
for(int i = 1; i <= N; i++) { //加入所有結點
int u = -1, min_cost = INF;
for(int j = 1; j <= N; j++) {
if(vis[j] == false && d[j] < min_cost) {//查詢距樹最近的結點
min_cost = d[j];
u = j;
}
}
if(u == -1) return -1;//非連通圖,構造MST失敗 可是WA啊
vis[u] = true; //標記訪問
ans += d[u]; //累加權值
for(int v = 1; v <= N; v++) { //更新最短距離
if(vis[v] == false && G[u][v] < d[v]){
d[v] = G[u][v];
}
}//for - v
}//for - i
return ans;
}//prime()
int main() {
int a, b, cost;
while(scanf("%d %d", &M, &N) != EOF) {
if(M == 0) break;
fill(G[0], G[0] + maxn * maxn, INF);
for(int i = 0; i < M; i++) {
scanf("%d %d %d", &a, &b, &cost);
G[a][b] = cost;
G[b][a] = cost;
}
int ans = prime(1); //從1號結點出發尋找
if(ans == -1) printf("?\n");
else printf("%d\n", ans);
}//while
system("pause");
return 0;
}
| Prime
演算法 - 鄰接表版本
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
const int maxn = 110;
const int INF = 0x3fffffff;
int d[maxn];
int vis[maxn];
struct edge {
int v, cost; //end, cost
edge() {}
edge(int _v, int _cost) : v(_v), cost(_cost) {} //建構函式,方便加入結點
};
vector<edge> Adj[maxn]; //Adjacency list
int N, M;
int prime(int st) {
fill(d, d + maxn, INF);
memset(vis, false, sizeof(vis));
d[st] = 0;//start
int ans = 0;
for(int i = 1; i <= N; i++) { // add all N nodes
int u = - 1, min_cost = INF;
for(int j = 1; j <= N; j++) {
if(vis[j] == false && d[j] < min_cost) {
min_cost = d[j];
u = j;
}
}
if(u == -1) return -1;
vis[u] = true; //add this node
ans += d[u]; //累加權值
for(int j = 0; j < Adj[u].size(); j++) {
int v = Adj[u][j].v, cost = Adj[u][j].cost;
if(vis[v] == false && cost < d[v])
d[v] = cost;
}
}//for - i;
return ans;
}
int main() {
int st, ed, cost;
edge e;
while(scanf("%d %d", &M, &N) != EOF) {
if(M == 0) break;
for(int i = 1; i <= N; i++) Adj[i].clear();
for(int i = 0; i < M; i++) {//input info of edges
scanf("%d %d %d", &st, &ed, &cost);
Adj[st].push_back(edge(ed, cost)); //underected graph
Adj[ed].push_back(edge(st, cost));
}
int ans = prime(1);//start from node 1
if(ans == -1) printf("?\n");
else printf("%d\n", ans);
}
system("pause");
return 0;
}