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先求出原图的最小生成树,然后枚举买哪些套餐,把一个套餐内的点相当与边权为0,直接用并查集缩点。正确性是基于一个贪心,
在Kruskal中,对于没有进入最小生成树的边,排序在它前面的边不会减少。
边比较多,用prim求最小生成树,效果比Kruskal好,枚举套餐的时候在用Kruskal。
#include<bits/stdc++.h> using namespace std; const int maxn = 1005; int n,q; int C[9]; vector<int> Buy[9]; #define PB push_back int x[maxn],y[maxn]; #define squ(x) ((x)*(x)) int dist(int a,int b) { return squ(x[a]-x[b])+squ(y[a]-y[b]); } struct Edge { int u,v,w; Edge(){} Edge(int u,int v,int w):u(u),v(v),w(w){} bool operator < (const Edge& x) const { return w > x.w; } }edges[maxn]; bool EdgeLess(const Edge &x,const Edge &y) { return x.w < y.w; } int ecnt; int d[maxn]; bool done[maxn]; const int INF = 0x3f3f3f3f; int Prim() { fill(d,d+n,INF); fill(done,done+n,0); ecnt = 0; priority_queue<Edge> q; q.push(Edge(-1,0,0)); // dummy edge int tot = d[0] = 0; while(q.size()){ Edge x = q.top(); q.pop(); if(done[x.v]) continue; edges[ecnt++] = x; tot += x.w; done[x.v] = true; for(int i = 1; i < n; i++){ if(done[i]) continue; int cost = dist(x.v,i); if(d[i]>cost){ d[i] = cost; q.push(Edge(x.v,i,cost)); } } } return tot; } int pa[maxn]; int Find(int x) { return x==pa[x]?x:pa[x]=Find(pa[x]); } void Union(int a,int b,int &cnt) { int s1 = Find(a),s2 = Find(b); if(s1 != s2){ pa[s1] = s2; cnt--; } } int Kruskal(int cnt) { if(!cnt) return 0; int ans = 0; for(int i = 1; i < ecnt; i++){ Edge &e = edges[i]; int s1 = Find(e.u), s2 = Find(e.v); if(s1 != s2) { ans += e.w; pa[s1] = s2; cnt--; if(!cnt) return ans; } } return ans; } int main() { //freopen("in.txt","r",stdin); int T; scanf("%d",&T); while(T--){ scanf("%d%d",&n,&q); for(int i = 0; i < q; i++){ int t; scanf("%d%d",&t,C+i); Buy[i].clear(); while(t--) { int c; scanf("%d",&c); Buy[i].PB(c-1); } } for(int i = 0; i < n; i++){ scanf("%d%d",x+i,y+i); } int ans = Prim(); sort(edges+1,edges+ecnt,EdgeLess); for(int mask = 1,M = 1<<q; mask < M; mask++){ for(int i = 0; i < n; i++) pa[i] = i; int tot = 0,cnt = n-1; for(int i = 0; i < q; i++){ if(mask&1<<i){ tot += C[i]; for(int j = 1; j < Buy[i].size(); j++) { Union(Buy[i][0],Buy[i][j],cnt); } } } tot += Kruskal(cnt); ans = min(ans,tot); } printf("%d\n",ans); if(T) putchar(‘\n‘); } return 0; }
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原文地址:http://www.cnblogs.com/jerryRey/p/4758297.html