題目鏈接

題意

N個商店最短修多長的路使得每個商店相連，其中apple和nike必須直接相連。

AC

• kruskal求最短路

#include <iostream> #include <stdio.h> #include <vector> #include <map> #include <string.h> #define mem(a, b) memset(a, b, sizeof(a)) #define ll long long #include <cmath> #include <algorithm> #define N 100005 #define P pair<int,int> #define mk(a, b) make_pair(a, b) int inf = 0x3f3f3f3f; using namespace std; struct ac{double len;int s, e; }b[3000]; struct point{int x, y; }a[100]; int pre[100]; void init() {for(int i = 0; i < 100; ++i) {pre[i] = i;} } int find(int x) {if (x != pre[x]) {return pre[x] = find(pre[x]);}return pre[x]; } void join(int x, int y) {if (x < y) pre[y] = x, find(y);else pre[x] = y, find(x);for (int i = 0; i < 100; ++i) {find(i);} } int main(){// freopen("in.txt", "r", stdin);int nike, apple;int n;while (scanf("%d", &n), n){scanf("%d%d", &nike, &apple);for (int i = 0; i < n; ++i) {scanf("%d%d", &a[i].x, &a[i].y);}init();double ans = 0;int now = 0;for (int i = 0; i < n; ++i) {for (int j = i + 1; j < n; ++j) {double len = sqrt((a[i].x - a[j].x) * (a[i].x - a[j].x) + (a[i].y - a[j].y) * (a[i].y - a[j].y));b[now].s = i;b[now].e = j;b[now].len = len;if (i == nike - 1 && j == apple - 1) {ans += b[now].len;// 將 nike 和 apple 相連join(i, j);}now++;}}sort(b, b + now, [&](const ac &a, const ac & b){if ((a.len - b.len) > 1e-6) {return 0;}else {return 1;}});for (int i = 0; i < now; ++i){int s = b[i].s, e = b[i].e;if (find(s) != find(e)) {// 將老大連起來join(find(s), find(e));ans += b[i].len;}}printf("%.2lf\n", ans);}return 0; }

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