蓝桥杯官网 试题 PREV-230 历届真题 作物杂交【第十一届】【决赛】【研究生组】【C++】【C】【Java】【Python】四种解法

为帮助大家能在6月18日的比赛中有一个更好的成绩，我会将蓝桥杯官网上的历届决赛题目的四类语言题解都发出来。希望能对大家的成绩有所帮助。

今年的最大目标就是能为【一亿技术人】创造更高的价值。

资源限制

内存限制：256.0MB   C/C++时间限制：1.0s   Java时间限制：3.0s   Python时间限制：5.0s

C++

//#pragma GCC optimize(3)
//#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
//#include
#include
//#include
#include
using namespace std;
#define ll    long long
#define ld    long double
#define ull   unsigned long long
#define ri    register int
#define rll   register long long
#define INF   0x7f7f7f7f
#define rep(i,begin,end)  for(int i=begin;i<=end;++i)
#define REP(i,begin,end)  for(int i=begin;i>=end;--i)
#define vi    vector
#define vl    vector
#define umap  unordered_map
#define uset  unordered_set
#define pii   pair
#define pll   pair
#define all(x)  x.begin(),x.end()
#define mkp   make_pair
#define y1    __y1
#define NL    putchar('\n')
#define MSET(a,v) memset(a,v,sizeof(a))
#define MT    int __T=RD;while(__T--)
#define BEFORE(__time)  while((double)clock()/CLOCKS_PER_SEC<__time)
inline int    min(int a,int b,int c){return min(a,min(b,c));}
inline ll    min(ll a,ll b,ll c){return min(a,min(b,c));}
inline double  min(double a,double b,double c){return min(a,min(b,c));}
inline int    max(int a,int b,int c){return max(a,max(b,c));}
inline ll    max(ll a,ll b,ll c){return max(a,max(b,c));}
inline double  max(double a,double b,double c){return max(a,max(b,c));}
inline int    gcd(int a,int b){int aa=a,bb=b;while(bb){int r=aa%bb;aa=bb;bb=r;}return aa;}
inline ll    gcd(ll a,ll b){ll aa=a,bb=b;while(bb){ll r=aa%bb;aa=bb;bb=r;}return aa;}
inline int    lcm(int a,int b){return a/gcd(a,b)*b;}
inline ll    lcm(ll a,ll b){return a/gcd(a,b)*b;}
inline int    lowbit(int x){return x&(-x);}
inline void    print(int x){printf("%d ",x);}
inline void    print(ll x){printf("%lld ",x);}
inline void    print(double x){printf("%lf ",x);}
inline void    print(int* a,int n)    {rep(i,1,n)printf("%d ",a[i]);NL;}
inline void    print(ll* a,int n)    {rep(i,1,n)printf("%lld ",a[i]);NL;}
inline void    print(double* a,int n)  {rep(i,1,n)printf("%.2lf ",a[i]);NL;}
/*
inline void   print(vi& a)      {for(auto&v:a)printf("%d ",v);NL;}
inline void   print(vl& a)      {for(auto&v:a)printf("%lld ",v);NL;}
inline void   print(vector& a){for(auto&v:a)printf("%.2lf ",v);NL;}
inline void   print(map& a)  {for(auto it=a.begin(),end=a.end();it!=end;++it)printf("%d %d\n",it->first,it->second);NL;}
inline void   print(umap& a) {for(auto it=a.begin(),end=a.end();it!=end;++it)printf("%d %d\n",it->first,it->second);NL;}
inline void   print(set& a)    {for(auto it=a.begin(),end=a.end();it!=end;++it)printf("%d ",it);NL;}
inline void   print(uset& a)   {for(auto it=a.begin(),end=a.end();it!=end;++it)printf("%d ",it);NL;}
*/
#define PR    print
inline int    read1(){ri  x=0;ri c=getchar();while(c<'0'||c>'9')c=getchar();while(c>='0'&&c<='9')x=x*10+(c^48),c=getchar();return x;}
inline ll    read2(){rll x=0;ri c=getchar();while(c<'0'||c>'9')c=getchar();while(c>='0'&&c<='9')x=x*10L+(c^48),c=getchar();return x;}
inline int    read3(){ri  x=0;ri c=getchar();int f=1;while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}while(c>='0'&&c<='9')x=x*10+(c^48),c=getchar();return x*f;}
inline ll    read4(){rll x=0;ri c=getchar();ll f=1L;while(c<'0'||c>'9'){if(c=='-')f=-1L;c=getchar();}while(c>='0'&&c<='9')x=x*10L+(c^48),c=getchar();return x*f;}
#define RD    read1()
#define _DIR_COUNT 0
#if (_DIR_COUNT==4)
const int dx[]={-1, 1, 0, 0}; //上下左右
const int dy[]={ 0, 0,-1, 1};
#elif (_DIR_COUNT==8)
const int dx[]={-1,-1, 0, 1, 1, 1, 0,-1}; //从上方顺时针
const int dy[]={ 0, 1, 1, 1, 0,-1,-1,-1};
#endif

const int N=2002;
int n,m,k,t;
int tim[N],type[N];
int f[N];
vector<pii>e[N];
void dfs(int p){
  if(f[p]!=-1)return;
  f[p]=0;
  int sz=e[p].size();
  int mi=INT_MAX;
  for(int i=0;i<sz;++i){
    int a=e[p][i].first,b=e[p][i].second;
    dfs(a);dfs(b);
    mi=min(mi,
      max(f[a],f[b])+max(tim[a],tim[b])
    );
  }
  if(mi!=INT_MAX)f[p]+=mi;
}
int main(){
  memset(f,-1,sizeof(f));
  
  n=RD;m=RD;k=RD;t=RD;
  rep(i,1,n)tim[i]=RD;
  rep(i,1,m)type[i]=RD;
  rep(i,1,k){
    int a=RD,b=RD,c=RD;
    e[c].push_back({a,b});
  }
  dfs(t);
  printf("%d",f[t]);
  return 0;
}

C

#include 

int n=0;//作物类型总数
int m=0;//拥有作物种数 
int k=0;//杂交方案数 
int t=0;// 目标种子编号

int N[2005]={0};//每种作物杂交天数
int M[2005]={0};//表示已有的种子类型 1 有   0 没有 
int K[100005][3]={0};//方案 

int f[2005];

int max(int a, int b){
  return a>b?a:b;
}
int min(int a, int b){
  return a}

void F( int sign ){

  int i=0;
  int minv = 1e9;
    
  if( f[sign] != -1 ){
    return ;
  }
  f[sign] = 0;
  
  for( i=1; i<=k; i++ ){
    if( sign == K[i][2] ){
      F( K[i][0] );
      F( K[i][1] );
      minv =  min(minv, max(f[ K[i][0]] , f[ K[i][1] ]) + max(N[ K[i][0] ], N[ K[i][1] ]) );
    }
    
  }
  if(minv != 1e9) f[sign] += minv;
  
} 
/*
6 2 4 6
5 3 4 6 4 9
1 2
1 2 3
1 3 4
2 3 5
4 5 6

*/

int main(){
  
  int i=0;
  memset(f,-1,sizeof f);
  scanf("%d %d %d %d",&n,&m,&k,&t);
  for( i=1; i<=n; i++ ){
    scanf("%d",&N[i]);
  }
  for( i=1; i<=m; i++ ){
    scanf("%d",&M[i]);
  }
  for( i=1; i<=k; i++ ){  // 0 X 1 == 2
    scanf("%d %d %d",&K[i][0],&K[i][1],&K[i][2]);
  }
  
  
  //查找t所需作物 
  F(t);
  
  
  printf("%d",f[t]);
  return 0;
}

Java

import java.awt.Point;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
 
public class Main {
  static int aa[], dp[];
  static List<Point> bh[];
 
  public static void main(String[] args) throws IOException {
    StreamTokenizer x = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
    PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
    x.nextToken();
    int n = (int) x.nval;
    x.nextToken();
    int m = (int) x.nval;
    x.nextToken();
    int k = (int) x.nval;
    x.nextToken();
    int t = (int) x.nval;
    aa = new int[n + 1];
    dp = new int[n + 1];
    bh = new List[n + 1];
    Arrays.fill(dp, Integer.MAX_VALUE);
    int bb[] = new int[m + 1];
    for (int i = 1; i <= n; i++) {
      x.nextToken();
      aa[i] = (int) x.nval;
      bh[i] = new ArrayList<Point>();
    }
    for (int i = 1; i <= m; i++) {
      x.nextToken();
      bb[i] = (int) x.nval;
      dp[bb[i]] = 0;
    }
    for (int i = 0; i < k; i++) {
      x.nextToken();
      int xx = (int) x.nval;
      x.nextToken();
      int yy = (int) x.nval;
      x.nextToken();
      int nn = (int) x.nval;
      bh[nn].add(new Point(xx, yy));
    }
    out.println(dfs(t));
    out.flush();
  }
 
  public static int dfs(int n) {
    if (dp[n] != Integer.MAX_VALUE)
      return dp[n];
    List<Point> p = bh[n];
    for (int i = 0; i < p.size(); i++) {
      int x = p.get(i).x, y = p.get(i).y;
      dp[n] = Math.min(dp[n], Math.max(dfs(x), dfs(y)) + Math.max(aa[x], aa[y]));
    }
    return dp[n];
  }
}

Python

def dfs(u):
    if f[u]:return f[u]
    L=len(s[u])
    if L==0:return 0
    f[u]=int(2e9)
    for i in range(0,L,2):
        f[u]=min(f[u],max(dfs(s[u][i]),dfs(s[u][i+1]))+max(t[s[u][i]-1],t[s[u][i+1]-1]))
    return f[u]
n,m,k,tar=map(int,input().split())
t=list(map(int,input().split()))
s=[]
f=[]
for i in range(n+1):
    s.append([])
    f.append(0)
a=list(map(int,input().split()))
while k:
    k-=1
    A,B,C=map(int,input().split())
    s[C].append(A)
    s[C].append(B)
print(dfs(tar))