HDU 3405 World Islands（最小生成树）

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水题。

少一个点的最小生成树，数据很小，n < 50，扫描n个点进行n次prim就好了。

代码：

// Header.#include <algorithm>#include <iostream>#include <sstream>#include <cstring>#include <cstdio>#include <vector>#include <string>#include <bitset>#include <queue>#include <stack>#include <cmath>#include <ctime>#include <set>#include <map>using namespace std;// Macrotypedef long long LL;#define TIME cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s." << endl;#define IN freopen("/Users/apple/input.txt", "r", stdin);#define OUT freopen("/Users/apple/out.txt", "w", stdout);#define mem(a, n) memset(a, n, sizeof(a))#define rep(i, n) for(int i = 0; i < (n); i ++)#define REP(i, t, n) for(int i = (t); i < (n); i ++)#define FOR(i, t, n) for(int i = (t); i <= (n); i ++)#define ALL(v) v.begin(), v.end()#define Min(a, b) a = min(a, b)#define Max(a, b) a = max(a, b)#define put(a) printf("%d\n", a)#define ss(a) scanf("%s", a)#define si(a) scanf("%d", &a)#define sii(a, b) scanf("%d%d", &a, &b)#define siii(a, b, c) scanf("%d%d%d", &a, &b, &c)#define VI vector<int>#define pb push_backconst int inf = 0x3f3f3f3f, N = 5e1 + 5, MOD = 1e9 + 7;// Macro endint T, cas = 0;int n, m;int vis[N], flag[N];double dis[N], g[N][N], ans;struct point{int x, y;}p[N];struct Edge{int u, v;double dis;bool operator < (const Edge& rhs) const {return dis < rhs.dis;}}e[N * N];// Impdouble Dis(int i, int j) {return sqrt((double)(p[i].x - p[j].x) * (p[i].x - p[j].x) + (double)(p[i].y - p[j].y) * (p[i].y - p[j].y));}void Prim() {mem(vis, 0);int u = flag[0];rep(i, n) if(!flag[i]) dis[i] = g[u][i];double sum = 0.0;vis[u] = 1;double minc;int p;REP(i, 1, n - 1) {minc = inf;rep(j, n) {if(!vis[j] && !flag[j] && dis[j] < minc) {minc = dis[j];p = j;}}sum += minc;vis[p] = 1;rep(j, n) if(!flag[j]) Min(dis[j], g[p][j]);}Min(ans, sum);}int main(){#ifdef LOCAL    IN;//  OUT;#endif    si(T);    while(T --) {    si(n);    rep(i, n) sii(p[i].x, p[i].y);    rep(i, n) rep(j, n) g[i][j] = g[j][i] = Dis(i, j);    mem(flag, 0);    ans = inf;    flag[0] = 1;    Prim();    REP(i, 1, n) {    flag[i-1] = 0, flag[i] = 1;    Prim();    }    printf("%.2lf\n", ans);    }    return 0;}

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