ICPC North Central NA Contest 2017 部分题解
大意：用最短路径来抓住所有的稀有精灵，DFS求最短路
#include <bits/stdc++.h>
#define mem(a) memset(a,0,sizeof(a))
#define forn(i,n) for(int i=0;i<n;++i)
#define for1(i,n) for(int i=1;i<=n;++i)
#define IO std::ios::sync_with_stdio(false); std::cin.tie(0)
#define ll long long
using namespace std;
const int maxn=22;
int n,m,all_mask;
int x[maxn],y[maxn],dist[maxn][maxn];
string name[maxn];
int dp[maxn][1<<maxn],p[maxn];
int dfs(int k,int mask)
{
if(dp[k][mask]!=-1) return dp[k][mask];
if(mask==all_mask) return dist[k][n];
int &ans=dp[k][mask];
ans=1e9;
forn(i,n)
{
if((1<<i)& mask) continue;
ans=min(ans,dist[k][i]+dfs(i,mask|p[i]));
}
return ans;
}
int main()
{
cin>>n;
forn(i,n)
cin>>x[i]>>y[i]>>name[i];
forn(i,n)
forn(j,n)
if(name[i]==name[j])
p[i]|=1<<j;
x[n]=y[n]=0;
p[n]=1<<n;
all_mask=(1<<n)-1;
for(int i=0;i<=n;++i)
for(int j=0;j<=n;++j)
dist[i][j]=abs(x[i]-x[j])+abs(y[i]-y[j]);
memset(dp,-1,sizeof(dp));
cout<<dfs(n,0)<<endl;
return 0;
}
C. Urban Design
解题过程：只要使用二分定理，只要判断两个点在哪些线的一侧，根据线的个数的奇偶来判断是否在相同的一块区域之中
#include <bits/stdc++.h>
#define mem(a) memset(a,0,sizeof(a))
#define forn(i,n) for(int i=0;i<n;++i)
#define for1(i,n) for(int i=1;i<=n;++i)
#define IO std::ios::sync_with_stdio(false); std::cin.tie(0)
#define ll long long
#define PI 3.14159265358979323846
using namespace std;
const int maxn=2e4+50;
int n,m,sx,sy,ans;
struct vec{int x1,x2,y1,y2,ans;}a[maxn],b[maxn];
ll cross(int x1,int y1,int x2,int y2){return 1LL*x1*y2-1LL*x2*y1;}
bool PointAndSide(const vec& line,const vec& pts)
{
int ox=line.x1,oy=line.y1;
ll dx=line.x2-line.x1,dy=line.y2-line.y1;
ll a=cross(pts.x1-ox,pts.y1-oy,dx,dy);
ll b=cross(pts.x2-ox,pts.y2-oy,dx,dy);
assert(a!=0&&b!=0);
return (a<0&&b>0)||(a>0&&b<0);
}
int main()
{
cin>>n;
forn(i,n)
cin>>a[i].x1>>a[i].y1>>a[i].x2>>a[i].y2;
cin>>m;
forn(i,m)
cin>>b[i].x1>>b[i].y1>>b[i].x2>>b[i].y2;
forn(i,n)
forn(j,m)
if(PointAndSide(a[i],b[j]))
b[j].ans++;
forn(i,m)
if(b[i].ans%2)
cout<<"different"<<endl;
else
cout<<"same"<<endl;
return 0;
}
G. Sheba‘s Amoebas
解题思路：没啥幺蛾子，直接DFS爆搜，直接过，数据量也不大
#include <bits/stdc++.h>
#define mem(a) memset(a,0,sizeof(a))
#define forn(i,n) for(int i=0;i<n;++i)
#define for1(i,n) for(int i=1;i<=n;++i)
#define IO std::ios::sync_with_stdio(false); std::cin.tie(0)
#define ll long long
using namespace std;
const int maxn=1e3+50;
char pic[110][110];
int dir[8][2] = { {1, 0}, {0, 1}, {-1, 0}, {0, -1}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1} };
bool vis1[110][110],vis2[110][110];
int n,m,sx,sy,ans;
bool check(int x, int y)
{
if(x >= 1 && x <= n & y >= 1 && y <= m && pic[x][y] == ‘#‘ && !vis1[x][y] & ! vis2[x][y])
return true;
return false;
}
void dfs(int x,int y,int cnt)
{
bool flag=false;
for(int i=0; i<8; ++i)
{
int tx=x+dir[i][0],ty=y+dir[i][1];
if(check(tx,ty))
{
vis1[tx][ty]=vis2[tx][ty]=true;
dfs(tx,ty,cnt+1);
vis1[tx][ty]=0;
flag=true;
}
}
if(!flag&&cnt>1)
{
for(int i = 0; i <= 7; i++)
{
int tx = x + dir[i][0], ty = y + dir[i][1];
if(tx == sx && ty == sy)
{
ans++;
return;
}
}
}
}
int main()
{
while(cin>>n>>m)
{
mem(vis1);
mem(vis2);
for1(i,n)
for1(j,m)
cin>>pic[i][j];
ans=0;
for1(i,n)
for1(j,m)
{
if(!vis1[i][j]&&pic[i][j]==‘#‘)
{
mem(vis1);
vis1[i][j]=vis2[i][j]=1;
sx=i,sy=j;
dfs(i,j,1);
}
}
cout<<ans<<endl;
}
return 0;
}
H. Zebras and Ocelots
解题思路：将整个排序反过来，然后看O的位置，O在第i位就会产生 2^i 的步骤，用快速幂加一下就OK
#include <bits/stdc++.h>
#define mem(a) memset(a,0,sizeof(a))
#define forn(i,n) for(int i=0;i<n;++i)
#define for1(i,n) for(int i=1;i<=n;++i)
#define IO std::ios::sync_with_stdio(false); std::cin.tie(0)
#define ll long long
using namespace std;
const int maxn=1e3+50;
int n,m,sx,sy,ans;
char c;
ll quick_pow(ll a,ll b)
{
ll res=1;
while(b)
{
if(b&1) res*=a;
a=a*a;
b=b>>1;
}
return res;
}
int main()
{
while(cin>>n)
{
ll ans=0;
string str;
forn(i,n)
{
cin>>c;
str+=c;
}
//cout<<str<<endl;
reverse(str.begin(),str.end());
for(int i=0;i<str.size();++i)
{
if(str[i]==‘O‘)
ans+=quick_pow(2,i);
}
cout<<ans<<endl;
}
return 0;
}
I. Racing Around the Alphabet
解题思路：纯模拟题，就是依次走，计算最短路径和就行了
#include <bits/stdc++.h>
#define mem(a) memset(a,0,sizeof(a))
#define forn(i,n) for(int i=0;i<n;++i)
#define for1(i,n) for(int i=1;i<=n;++i)
#define IO std::ios::sync_with_stdio(false); std::cin.tie(0)
#define ll long long
#define PI 3.14159265358979323846
using namespace std;
const int maxn=1e3+50;
int n,m,sx,sy,ans;
double perlen,res;
char lists[28]={‘ ‘,‘\‘‘,‘A‘,‘B‘,‘C‘,‘D‘,‘E‘,‘F‘,‘G‘,‘H‘,‘I‘,‘J‘,‘K‘,‘L‘,‘M‘,‘N‘,‘O‘,‘P‘,‘Q‘,‘R‘,‘S‘,‘T‘,‘U‘,‘V‘,‘W‘,‘X‘,‘Y‘,‘Z‘};
char dim[200];
int searchx(char x)
{
for(int i=0;i<28;++i)
{
if(lists[i]==x)
return i+1;
}
}
int main()
{
perlen=PI*60.0/28.0;
while(cin>>n)
{
getchar();
while(n--)
{
res=0.0;
string str;
getline(cin,str);
res+=str.size();
for(int i=0;i<str.size()-1;++i)
{
int a=searchx(str[i])-searchx(str[i+1]);
if(a<0)
a=min(abs(a),abs(a+28));
else
a=min(a,28-a);
//cout<<a<<endl;
res+=perlen*a/15.0;
}
printf("%.10f\n",res);
}
}
return 0;
}
J. Lost Map
解题思路：最小生成树，模板题，随便套Kruskal和Prim随便用一个就行了
#include <bits/stdc++.h>
using namespace std;
#define MAX_N 2501
struct edge {
int from,to;
long long cost;
}E[MAX_N*MAX_N];
int N,M;//N是点的个数，M是边的数量
int father[MAX_N];
void init()
{
for(int i=1;i<=N;i++)father[i]=i;
}
int findd(int x)
{
if(x==father[x])return x;
return father[x]=findd(father[x]);
}
bool same(int x,int y)
{
return findd(x)==findd(y);
}
void unite(int x,int y)
{
int u=findd(x),v=findd(y);
if(u==v)return;
father[u]=v;
return;
}
bool cmp(edge a,edge b)
{
return a.cost<b.cost;
}
void Kruskal()
{
sort(E+1,E+1+M,cmp);
for(int i=1;i<=M;i++)
{
if(same(E[i].from,E[i].to))continue;
unite(E[i].from,E[i].to);
cout<<E[i].from<<" "<<E[i].to<<endl;
}
return;
}
int main()
{
int x;
while(~scanf("%d",&N))
{
M=0;
init();
for(int i=0;i<N;i++)
{
for(int j=0;j<N;j++)
{
scanf("%d",&x);
if(x!=0)
{
M++;
E[M].from=i+1;
E[M].to=j+1;
E[M].cost=x;
}
}
}
Kruskal();
}
return 0;
}
原文地址：https://www.cnblogs.com/bethebestone/p/12383619.html
时间： 2024-07-30 23:01:37