Buy or Build UVA - 1151 Kruskal+枚举-CFANZ编程社区

题意：

大概意思是有 n 个点，现在有 q 个方案，第 i 个方案耗费为 ci ，使 Ni 个点联通，当然也可以直接使两点联通，现求最小生成树的代价。

两点直接联通的代价是欧几里得距离的平方；

由于0<=q<=8，所以我们考虑二进制枚举；该位为1表示选择该方案，然后每次求一遍cost ，最后取 min 即可；

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
//#include
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 1000005
#define inf 0x3f3f3f3f
#define INF 9999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair pii;
#define pi acos(-1.0)
const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair pii;
inline ll rd() {
    ll x = 0;
    char c = getchar();
    bool f = false;
    while (!isdigit(c)) {
        if (c == '-') f = true;
        c = getchar();
    }
    while (isdigit(c)) {
        x = (x << 1) + (x << 3) + (c ^ 48);
        c = getchar();
    }
    return f ? -x : x;
}

ll gcd(ll a, ll b) {
    return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
    if (!b) {
        x = 1; y = 0; return a;
    }
    ans = exgcd(b, a%b, x, y);
    ll t = x; x = y; y = t - a / b * y;
    return ans;
}
*/



ll qpow(ll a, ll b, ll c) {
    ll ans = 1;
    a = a % c;
    while (b) {
        if (b % 2)ans = ans * a%c;
        b /= 2; a = a * a%c;
    }
    return ans;
}
int T;
int n, q;
int cost[maxn];

int fa[maxn];
int cnt;
vectorvc[maxn];

struct point {
    int x, y;
}pint[maxn];

struct node {
    int x, y;
    int w;
}edge[maxn];

bool cmp(node a, node b) {
    return a.w < b.w;
}

int dis(int a, int b, int x, int y) {
    return ((a - x)*(a - x) + (b - y)*(b - y));
}

void init(int n) {
    for (int i = 0; i <= n; i++)fa[i] = i;
}

int findfa(int x) {
    if (x == fa[x])return x;
    return fa[x] = findfa(fa[x]);
}

void Union(int p, int q) {
    if (findfa(q) != findfa(p)) {
        fa[findfa(p)] = findfa(q);
    }
}

int ans;

int kruskal() {
    int res = 0;
    int ct = 0;
    for (int i = 0; i < cnt; i++) {
        int u = edge[i].x; int v = edge[i].y;
        if (findfa(u) != findfa(v)) {
            Union(u, v); res += edge[i].w;
            ct++;
            if (ct == n - 1)break;
        }
    }
    return res;
}


void sol() {
    init(n);
    ans = kruskal();
    
    for (int i = 0; i < (1 << q); i++) {
        init(n);int cst = 0;
        for (int j = 0; j < q; j++) {
            if (((i >> j) & 1)==0)continue;
            cst += cost[j];
            for (int k = 1; k < vc[j].size(); k++) {
                Union(vc[j][k], vc[j][0]);
            }
        }
        ans = min(ans, cst + kruskal());
    }
    printf("%d\n", ans);
}

int main()
{
    //ios::sync_with_stdio(0);
    rdint(T); int kase = 1;
    while (T--) {
        if (kase > 1)printf("\n");
        kase++;
        rdint(n); rdint(q); cnt = 0;
        for (int i = 0; i < 10; i++)vc[i].clear();
        for (int i = 0; i < q; i++) {
            int tmp; rdint(tmp); rdint(cost[i]);
            while (tmp--) {
                int x; rdint(x); vc[i].push_back(x);
            }
        }
        
        for (int i = 1; i <= n; i++) {
            rdint(pint[i].x); rdint(pint[i].y);
        }
        for (int i = 1; i <= n; i++) {
            for (int j = i + 1; j <= n; j++) {
                edge[cnt].x = i; edge[cnt].y = j; edge[cnt].w = dis(pint[i].x, pint[i].y, pint[j].x, pint[j].y); cnt++;
            }
        }
        sort(edge, edge + cnt, cmp);
        sol();
        
    }

    return 0;
}

EPFL - Fighting