标签：and namespace puts 排序 1.0 cat algo start eps
显然线性基可以满足题目中给出的条件。关键是如何使得魔力最大。
贪心策略是按魔力排序，将编号依次加入线性基，一个数如果和之前的一些数异或和为0就跳过他。
因为如果要把这个数放进去，那就要把之前的某个数拿出来，而这样交换之后集合能异或出的数是不会变的，和却变小了。
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi acos(-1.0) # define eps 1e-3 # define MOD 1000000007 # define INF 1000000000 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; int Scan() { int res=0, flag=0; char ch; if((ch=getchar())==‘-‘) flag=1; else if(ch>=‘0‘&&ch<=‘9‘) res=ch-‘0‘; while((ch=getchar())>=‘0‘&&ch<=‘9‘) res=res*10+(ch-‘0‘); return flag?-res:res; } void Out(int a) { if(a<0) {putchar(‘-‘); a=-a;} if(a>=10) Out(a/10); putchar(a%10+‘0‘); } const int N=1005; //Code begin... typedef struct {LL num; int magic;}Node; Node a[N]; LL p[63]; bool comp(Node a, Node b){return a.magic>b.magic;} int main () { int n, ans=0; scanf("%d",&n); FOR(i,1,n) scanf("%lld%d",&a[i].num,&a[i].magic); sort(a+1,a+n+1,comp); FOR(i,1,n) { for (int j=62; j>=0; --j) { if (!(a[i].num>>j)) continue; if (!p[j]) {p[j]=a[i].num; ans+=a[i].magic; break;} a[i].num^=p[j]; } } printf("%d\n",ans); return 0; }
标签：and namespace puts 排序 1.0 cat algo start eps
原文地址：http://www.cnblogs.com/lishiyao/p/6372147.html
