noip2014题解

没有正解 都是我的暴力

T1生活大爆炸版石头剪刀布

把得分要求存进C数组里

c[i][j]表示i对j的得分情况

#include<iostream>
#include<cmath>
#include<ctime>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#define LL long long
#define N 202
using namespace std;

int n,na,nb;

int a[N],b[N],c[10][10];

int A,B;

void rule_()
{
    /*
    0--剪刀 1--石头 2--布  3--蜥蜴人  4--斯波克 
    */
    for(int i=0;i<=4;i++) c[i][i]=0;
    c[0][1]=0;c[0][2]=1;c[0][3]=1;c[0][4]=0;
    c[1][0]=1;c[1][2]=0;c[1][3]=1;c[1][4]=0;
    c[2][0]=0;c[2][1]=1;c[2][3]=0;c[2][4]=1;
    c[3][0]=0;c[3][1]=0;c[3][2]=1;c[3][4]=1;
    c[4][0]=1;c[4][1]=1;c[4][2]=0;c[4][3]=0;
}

void play_()
{
    for(int i=1;i<=n;i++)
    {
        int xa,xb;
        xa=i%na;if(!xa) xa=na;
        xb=i%nb;if(!xb) xb=nb;
        A=A+c[a[xa]][b[xb]];
        B=B+c[b[xb]][a[xa]]; 
    }
}

int main()
{
///    freopen("rps.in","r",stdin);
///    freopen("rps.out","w",stdout);
    scanf("%d%d%d",&n,&na,&nb);
    for(int i=1;i<=na;i++) scanf("%d",&a[i]);
    for(int i=1;i<=nb;i++) scanf("%d",&b[i]);
    rule_();
    play_();
    printf("%d %d\n",A,B);
//    fclose(stdin);
//    fclose(stdout);
    return 0;
}
/*
10 5 6
0 1 2 3 4
0 3 4 2 1 0

9 5 5
0 1 2 3 4
1 0 3 2 4
*/
100

T2联合权值

30% O(n^3)floyed求两点距离+n^2枚举

60% LCA求两点距离+n^2枚举

#include<iostream>
#include<cmath>
#include<ctime>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#define LL long long
#define N 2020
#define M 10007
using namespace std;

int n;

int sumedge;

int head[N],w[N];

int size[N],deep[N],dad[N],top[N];

int dis[103][103];

inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

struct Edge
{
    int x,y,nxt;
    Edge(int x=0,int y=0,int nxt=0):
        x(x),y(y),nxt(nxt){}
}edge[N<<1];

void add(int x,int y)
{
    edge[++sumedge]=Edge(x,y,head[x]);
    head[x]=sumedge;
}

void slove1()
{
    int max_,res;max_=res=0;
    memset(dis,0x3f,sizeof(dis));
    for(int i=1;i<=n;i++) dis[i][i]=0;
    for(int i=1;i<n;i++)
    {
        int x,y;
        x=read();y=read();
        dis[x][y]=dis[y][x]=1;
    }
    for(int k=1;k<=n;k++)
     for(int i=1;i<=n;i++)
      for(int j=1;j<=n;j++)
       dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);
    for(int i=1;i<=n;i++) w[i]=read();
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
    //        cout<<i<<" "<<j<<" "<<dis[i][j]<<endl;
            if(dis[i][j]!=2) continue;
    //        cout<<i<<" --- "<<j<<endl;
            int tmp;tmp=w[i]*w[j];
            max_=max(max_,tmp);
            res=(res%M+tmp%M)%M;
        }
    }
    printf("%d %d\n",max_,res);
}

void dfs(int x)
{
    size[x]=1;deep[x]=deep[dad[x]]+1;
    for(int i=head[x];i;i=edge[i].nxt)
    {
        int v=edge[i].y;
        if(v==dad[x]) continue;
        dad[v]=x;
        dfs(v);
        size[x]+=size[v];
    }
}

void dfs_(int x)
{
    int s=0;
    if(!top[x]) top[x]=x;
    for(int i=head[x];i;i=edge[i].nxt)
    {
        int v=edge[i].y;
        if(v!=dad[x]&&size[v]>size[s]) s=v;
    }
    if(s)
    {
        top[s]=top[x];
        dfs_(s);
    }
    for(int i=head[x];i;i=edge[i].nxt)
    {
        int v=edge[i].y;
        if(v!=dad[x]&&v!=s)dfs_(v);
    }
}

int LCA(int x,int y)
{
    if(deep[x]>deep[y]) swap(x,y);
    for(;top[x]!=top[y];)
    {
        if(deep[top[x]]>deep[top[y]]) swap(x,y);
        y=dad[top[y]];
    }
    if(deep[x]>deep[y]) swap(x,y);
    return x; 
}

void slove2()
{
    int max_,res;max_=res=0;
    for(int i=1;i<n;i++)
    {
        int x,y;
        x=read();y=read();
        add(x,y);add(y,x);
    }
    for(int i=1;i<=n;i++) w[i]=read();
    deep[0]=-1;
    dfs(1);dfs_(1);
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            int D,tmp;D=deep[i]+deep[j]-2*deep[LCA(i,j)];
            if(D!=2) continue;tmp=w[i]*w[j];    
            max_=max(max_,tmp);
            res=(res%M+tmp%M)%M;
        }
    }
    printf("%d %d\n",max_,res);
}

int main()
{
//    freopen("linkb.in","r",stdin);
//    freopen("linkb.out","w",stdout);
    n=read();
    if(n<=100) slove1();
    else 
    if(n<=2000) 
    slove2();
//    fclose(stdin);
//    fclose(stdout);
    return 0;
}
/*
5
1 2
2 3
3 4 
4 5
1 5 2 3 10

4
1 2
2 3
2 4
1 1 1 1
*/
60

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转载自www.cnblogs.com/zzyh/p/9918087.html