UVa1151&POJ2784--Buy or Build【kruskal+二进制枚举】

来源：互联网发布：二分查找java 编辑：程序博客网时间：2024/05/18 09:02

链接：

UVa http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3592

POJ http://poj.org/problem?id=2784

题意：告诉你n个点的坐标，建立一颗最小生成树，不过有q个套餐，套餐是连通某些点，并有一定花费，求最小生成树。

思路：n个点，n最大为1000，则最多有1000*999/2条边，先不使用套餐求一遍最小生成树，保留下最小生成树的边，此时最多有999条边，然后枚举每种套餐选或者不选，因为只有两种情况，用二进制枚举就行了，然后从筛选出的最多999条边里选边构造最小生成树。

#include<cstring>#include<string>#include<fstream>#include<iostream>#include<iomanip>#include<cstdio>#include<cctype>#include<algorithm>#include<queue>#include<map>#include<set>#include<vector>#include<stack>#include<ctime>#include<cstdlib>#include<functional>#include<cmath>using namespace std;#define PI acos(-1.0)#define MAXN 5000100#define eps 1e-7#define INF 0x7FFFFFFF#define LLINF 0x7FFFFFFFFFFFFFFF#define seed 131#define MOD 1000000007#define ll long long#define ull unsigned ll#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1struct node{    int u,v;    int w;}edge[MAXN],edge2[1010];int n,m,q,ans;int qc[10][1010],qn[10],qm[10];int x[1010],y[1010],father[1010];bool cmp(node x,node y){    return x.w<y.w;}int Find(int x){    int t = x;    while(t!=father[t])        t = father[t];    int k = x;    while(k!=t){        int temp = father[k];        father[k] = t;        k = temp;    }    return t;}void init(){    for(int i=0;i<=n;i++)   father[i] = i;}int kruskal(){    int i,j;    int sum = 0;    int tot = 0;    for(i=0;i<m;i++){        int a = Find(edge[i].u);        int b = Find(edge[i].v);        if(a!=b){            father[a] = b;            edge2[sum] = edge[i];            tot += edge[i].w;            sum++;            if(sum>=n-1)    break;        }    }    return tot;}int gao(int x){    int i,j;    int sum = 0;    int tot = 0;    for(i=0;i<q;i++){        if(x&(1<<i)){            tot += qm[i];            for(j=0;j<qn[i]-1;j++){                int aa = Find(qc[i][j]);                int bb = Find(qc[i][j+1]);                if(aa!=bb){                    father[aa] = bb;                    sum++;                }            }        }    }    for(i=0;i<n-1;i++){        int aa = Find(edge2[i].u);        int bb = Find(edge2[i].v);        if(aa!=bb){            father[aa] = bb;            sum++;            tot += edge2[i].w;            if(sum>=n-1)    break;        }    }    //cout<<tot<<endl;    return tot;}int main(){    int t,i,j;    scanf("%d",&t);    for(int iii=0;iii<t;iii++){        if(iii) puts("");        m = 0;        scanf("%d%d",&n,&q);        for(i=0;i<q;i++){            scanf("%d%d",&qn[i],&qm[i]);            for(j=0;j<qn[i];j++){                scanf("%d",&qc[i][j]);            }        }        for(i=1;i<=n;i++){            scanf("%d%d",&x[i],&y[i]);            for(j=1;j<i;j++){                int temp = (x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]);                edge[m].u = i;                edge[m].v = j;                edge[m].w = temp;                m++;            }        }        sort(edge,edge+m,cmp);        init();        ans = kruskal();        for(i=0;i<(1<<q);i++){            init();            int temp = gao(i);            ans = min(ans,temp);        }        printf("%d\n",ans);    }    return 0;}

0 0