版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/yixin94/article/details/82053897
题意：有一组卡片，正反面的值是R[i],B[i]，每次选择两个卡片i,j把min(R[i]^B[j],B[i]^R[j])加到value中，再将其中一个卡片放回，直到只剩一张卡片。问最优策略下最小的value是多少。
small input可以用dp，状态用二进制数表示，比如N=5,10101表示余下第0,2,4个卡片，状态转移时枚举选取其中两个卡片，再放回去一个。
large input真心想不到最小生成树上面去。切入点是假设最优解中的操作次序是选择卡片a,b,移除卡片b，那么说明a把b移除了，接下来选择卡片a,c,移除卡片a，a又被c移除，不可能出现的情况就是同时选了b,c，b把c移除或者c把b移除。Then 如果卡片之间连一条边，最优解中不可能出现环。因此就是个求最小生成树的问题。
卡片i,j之间连一条无向边，weight=min(R[i]^B[j],B[i]^R[j])。再求最小生成树即可。
#include<iostream>
#include<stdio.h>
#include<cstdio>
#include<string>
#include<cmath>
#include<stdlib.h>
#include<algorithm>
#include<string.h>
#include<cstring>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<windows.h>
using namespace std;
//Kcikstart 2017 Round D Problem B. Cards Game
const int maxn=110;
int T;
int N;
long long R[maxn];
long long B[maxn];
long long mp[maxn][maxn];
int p[maxn*maxn];// disjoint-sets forest
int E;
long long ans;
class edge
{
public:
int a;
int b;
long long weight;
edge()
{
a=0;
b=0;
weight=0;
}
edge(int a0,int b0,long long w0)
{
a=a0;
b=b0;
weight=w0;
}
bool operator < (edge other) const
{
return weight<other.weight;
}
bool operator > (edge other) const
{
return weight>other.weight;
}
bool operator == (edge other) const
{
return weight==other.weight;
}
};
edge edge_set[maxn*maxn];
int init()
{
for (int i=0;i<maxn*maxn;++i)
{
p[i]=i;
}
}
int find(int x)
{
if(x==p[x])
{
return x;
}
else
{
return p[x]=find(p[x]);
}
}
void Union(int x,int y)
{
p[find(x)]=find(y);
}
void Kruskal()
{
init();
sort(edge_set, edge_set+E);
// for(int i=0;i<E;i++)
// {
// cout<<edge_set[i].a<<" "<<edge_set[i].b<<" "<<edge_set[i].weight<<endl;
// }
int j=0;
while(j<E)
{
while(find(edge_set[j].a)==find(edge_set[j].b))
{
// cout<<edge_set[j].a<<" "<<edge_set[j].b<<" "<<p[edge_set[j].a]<<" "<<p[edge_set[j].b]<<endl;
j++;
if(j>=E)
{
return;
}
// cout<<j<<endl;
}
Union(edge_set[j].a,edge_set[j].b);
// cout<<"add edge "<<edge_set[j].a<<" "<<edge_set[j].b<<endl;
ans+=edge_set[j].weight;
j++;
}
}
int main()
{
// freopen("input.txt","r",stdin);
freopen("B-small-practice.in","r",stdin);
freopen("B.txt","w",stdout);
cin>>T;
for(int ca=1;ca<=T;ca++)
{
memset(R,0,sizeof(R));
memset(B,0,sizeof(B));
memset(mp,0,sizeof(mp));
memset(p,0,sizeof(p));
memset(edge_set,0,sizeof(edge_set));
scanf("%d",&N);
E=0;
ans=0;
for(int i=0;i<N;i++)
{
scanf("%lld",&R[i]);
}
for(int i=0;i<N;i++)
{
scanf("%lld",&B[i]);
}
// for(int i=0;i<N;i++)
// {
// cout<<R[i]<<" "<<B[i]<<endl;
// }
for(int i=0;i<N;i++)
{
for(int j=i+1;j<N;j++)
{
int weight=min(R[i]^B[j],R[j]^B[i]);
// cout<<weight<<" "<<(R[i]^B[j])<<" "<<(R[j]^B[i])<<endl;
mp[i][j]=weight;
mp[j][i]=weight;
edge_set[E++]=edge(i,j,weight);
}
}
// for(int i=0;i<N;i++)
// {
// for(int j=0;j<N;j++)
// {
// cout<<mp[i][j]<<" ";
// }
// cout<<endl;
// }
Kruskal();
printf("Case #%d: %lld\n",ca,ans);
}
return 0;
}
附dp code for small input
#include<iostream>
#include<stdio.h>
#include<cstdio>
#include<string>
#include<cmath>
#include<stdlib.h>
#include<algorithm>
#include<string.h>
#include<cstring>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<bitset>
#include<climits>
using namespace std;
//kickstart 2017 Round G Problem B
int T;
const int maxn=1<<6;
int N;
int M;
long long ans;
long long dp[maxn];
int used[maxn];
long long R[10];
long long B[10];
long long dfs(int state)
{
//cout<<"state "<<bitset<5>(state)<<endl;
if(used[state]==true)
{
return dp[state];
}
int cardnum=0;
int cardidx[N];
memset(cardidx,0,sizeof(cardidx));
for(int i=0;i<N;i++)
{
//cout<<"check "<<bitset<5>((1<<i))<<" "<<((1<<i)&state)<<endl;
if(((1<<i)&state)!=0)
{
cardnum++;
cardidx[i]=1;
//cout<<i<<endl;
}
}
if(cardnum==1)
{
dp[state]=0;
return 0;
}
int card1=0;
int card2=0;
long long minret=LLONG_MAX;
//bool chg=false;
for(int i=0;i<N;i++)
{
if(cardidx[i]==1)
{
card1=i;
for(int j=i+1;j<N;j++)
{
if(cardidx[j]==1)
{
card2=j;
long long curr=min(R[card1]^B[card2],R[card2]^B[card1]);
//cout<<curr<<" "<<minret<<" "<<(minret-curr)<<endl;
int nextstate=state^(1<<card1);
minret=min(minret,curr+dfs(nextstate));
nextstate=state^(1<<card2);
minret=min(minret,curr+dfs(nextstate));
//chg=true;
}
}
}
}
// if(chg==false)
// {
// cout<<"unchanged "<<bitset<5>(state)<<endl;
// }
//cout<<bitset<5>(state)<<" "<<minret<<endl;
dp[state]=minret;
used[state]=true;
return minret;
}
int main()
{
// long long sminret=LLONG_MAX;
// cout<<sminret<<endl;
//cout<<(872031131^582064085)<<" "<<0x3f3f3f3f<<endl;
freopen("B-small-attempt2.in","r",stdin);
//freopen("input.txt","r",stdin);
freopen("b1.txt","w",stdout);
scanf("%d",&T);
for(int ca=1;ca<=T;ca++)
{
ans=0;
memset(R,0,sizeof(R));
memset(B,0,sizeof(B));
memset(dp,0,sizeof(dp));
memset(used,false,sizeof(used));
scanf("%d",&N);
for(int i=0;i<N;i++)
{
dp[(1<<i)]=0;
used[(1<<i)]=true;
scanf("%lld",&R[i]);
}
for(int i=0;i<N;i++)
{
scanf("%lld",&B[i]);
}
//cout<<"init "<<((1<<N)-1)<<endl;
ans=dfs((1<<N)-1);
printf("Case #%d: %lld\n",ca,ans);
}
return 0;
}