codeforces#327 div2

这场状态不好有点可惜，题目都不难，而且很好。。

A题：水题。

#include<bits/stdc++.h>
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define MS0(a) memset(a,0,sizeof(a))
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

using namespace std;

typedef long long ll;
const int maxn=1000100;
const int INF=(1<<29);
const double EPS=0.000000000001;
const double Pi=acos(-1.0);

int l,p,q;

int main()
{
    //freopen("in.txt","r",stdin);
    while(cin>>l>>p>>q){
        printf("%.5f\n",l*1.0*p/(p+q));
    }
    return 0;
}

View Code

B题：水题。居然卡了，状态真差。。。

#include<bits/stdc++.h>
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define MS0(a) memset(a,0,sizeof(a))
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

using namespace std;

typedef long long ll;
const int maxn=200100;
const int INF=(1<<29);
const double EPS=0.000000000001;
const double Pi=acos(-1.0);

int n,q;
int val[26];
char s[maxn];
int len;
char u,v;
char op[maxn];

int main()
{
    //freopen("in.txt","r",stdin);
    while(cin>>n>>q){
        scanf("%s",s);
        len=strlen(s);
        REP(i,0,25) val[i]=i;
        gets(op);
        while(q--){
            //cin>>u>>v;
            gets(op);
            u=op[0],v=op[2];
            u-='a';v-='a';
            REP(i,0,25){
                if(val[i]==u) val[i]=v;
                else if(val[i]==v) val[i]=u;
            }
        }
        REP(i,0,len-1){
            s[i]=val[s[i]-'a']+'a';
        }
        printf("%s\n",s);
    }
    return 0;
}

View Code

C题：水题。这个有思路，但是想着点到分心了。。

#include<bits/stdc++.h>
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define MS0(a) memset(a,0,sizeof(a))
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

using namespace std;

typedef long long ll;
const int maxn=1000100;
const int INF=(1<<29);
const double EPS=0.000000000001;
const double Pi=acos(-1.0);

int n,a[maxn];

int main()
{
    //freopen("in.txt","r",stdin);
    while(cin>>n){
        REP(i,1,n) scanf("%d",&a[i]);
        int ans=0;
        for(int i=2;i<n;i++){
            if(a[i]^a[i-1]&&a[i]^a[i+1]){
                int l=i-1,r=i+1;
                while(l-1>=1&&a[l]^a[l-1]) l--;
                while(r+1<=n&&a[r]^a[r+1]) r++;
                //cout<<l<<" "<<r<<endl;
                if((r-l+1)&1){
                    REP(j,l,r) a[j]=a[r];
                    ans=max((r-l+1)/2,ans);
                }
                else{
                    int m=(l+r)/2;
                    REP(j,l,m) a[j]=a[l];
                    REP(j,m+1,r) a[j]=a[r];
                    ans=max((r-l+1)/2-1,ans);
                }
                i=r;
            }
        }
        cout<<ans<<endl;
        REP(i,1,n){
            if(i!=n) printf("%d ",a[i]);
            else printf("%d\n",a[i]);
        }
    }
    return 0;
}

View Code

D题：二分时间t。因为位移和运动轨迹并没有关系，位移也是满足矢量合成的，直接分解也可以。注意二分三分都写成for循环上1000次，不要带EPS，这样可以防止TLE。

#include<bits/stdc++.h>
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define MS0(a) memset(a,0,sizeof(a))
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

using namespace std;

typedef long long ll;
const int maxn=1000100;
const int INF=(1<<29);
const double EPS=0.0000000001;
const double Pi=acos(-1.0);

double X1,Y1,X2,Y2;
double vm,t1;
double vx,vy,wx,wy;

double dist(double x,double y,double x0,double y0)
{
    double tx=x-x0,ty=y-y0;
    return sqrt(tx*tx+ty*ty);
}

bool check(double t)
{
    if(t<t1+EPS){
        double x=X2-vx*t,y=Y2-vy*t;
        double d=dist(x,y,X1,Y1);
        //printf("%.1f %.1f %.1f %.1f %.1f %.1f %.1f \n",x,y,d,X1,Y1,X2,Y2);
        if(vm*t>d-EPS) return 1;
        return 0;
    }
    else{
        double x0=X2-vx*t1,y0=Y2-vy*t1;
        double x=x0-wx*(t-t1),y=y0-wy*(t-t1);
        double d=dist(x,y,X1,Y1);
        if(vm*t>d-EPS) return 1;
        return 0;
    }
}

double bin(double l,double r)
{
    double res=r;
    REP(i,1,1000){
        double m=(l+r)/2;
        if(check(m)) res=r=m;
        else l=m;
    }
    return res;
}

int main()
{
    //freopen("in.txt","r",stdin);
    while(cin>>X1>>Y1>>X2>>Y2>>vm>>t1>>vx>>vy>>wx>>wy){
        double t=bin(0,INF);
        printf("%.7f\n",t);
    }
    return 0;
}

View Code

E题：bfs预处理出每个点到三个国家的距离即可，然后路的长度就是三个距离直接加。

#include<bits/stdc++.h>
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define MS0(a) memset(a,0,sizeof(a))
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

using namespace std;

typedef long long ll;
const int maxn=1100;
const int INF=(1<<29);
const double EPS=0.0000001;
const double Pi=acos(-1.0);

int n,m;
char ch[maxn][maxn];
int d[3][maxn][maxn];
bool vis[maxn][maxn];
int dx[]={-1,1,0,0};
int dy[]={0,0,-1,1};
struct Node
{
    int x,y;
    int step;
};

void bfs(char c)
{
    memset(d[c-'1'],-1,sizeof(d[c-'1']));
    MS0(vis);
    queue<Node> q;
    REP(i,1,n){
        REP(j,1,m){
            if(ch[i][j]==c) q.push({i,j,0}),vis[i][j]=1;
        }
    }
    while(!q.empty()){
        Node u=q.front();
        q.pop();
        d[c-'1'][u.x][u.y]=u.step;
        REP(i,0,3){
            int nx=u.x+dx[i],ny=u.y+dy[i];
            if(ch[nx][ny]=='#'||vis[nx][ny]) continue;
            vis[nx][ny]=1;
            q.push({nx,ny,u.step+1});
        }
    }
}

void debug(char c)
{
    cout<<endl;
    REP(i,1,n){
        REP(j,1,m){
            printf("%2d",d[c][i][j]);
        }
        cout<<endl;
    }
}

void deal()
{
    int r1=INF,r2=INF;
    REP(i,1,n){
        REP(j,1,m){
            if(ch[i][j]=='1'){
                r1=min(d[1][i][j],r1);
                r2=min(d[2][i][j],r2);
            }
        }
    }
    REP(i,1,n) REP(j,1,m) if(ch[i][j]=='1') d[1][i][j]=r1,d[2][i][j]=r2;
    r1=INF,r2=INF;
    REP(i,1,n){
        REP(j,1,m){
            if(ch[i][j]=='2'){
                r1=min(d[0][i][j],r1);
                r2=min(d[2][i][j],r2);
            }
        }
    }
    REP(i,1,n) REP(j,1,m) if(ch[i][j]=='2') d[0][i][j]=r1,d[2][i][j]=r2;
    r1=INF,r2=INF;
    REP(i,1,n){
        REP(j,1,m){
            if(ch[i][j]=='3'){
                r1=min(d[0][i][j],r1);
                r2=min(d[1][i][j],r2);
            }
        }
    }
    REP(i,1,n) REP(j,1,m) if(ch[i][j]=='3') d[0][i][j]=r1,d[1][i][j]=r2;
}

int main()
{
   // freopen("in.txt","r",stdin);
    while(cin>>n>>m){
        memset(ch,'#',sizeof(ch));
        REP(i,1,n){
            REP(j,1,m) cin>>ch[i][j];
        }
        bfs('1');bfs('2');bfs('3');
        deal();
        //debug(0);debug(1);debug(2);
        int ans=INF;
        REP(i,1,n){
            REP(j,1,m){
                bool tag=1;
                REP(k,0,2){
                    if(d[k][i][j]==-1) tag=0;
                }
                if(!tag) continue;
                int tmp=0;
                tmp=d[0][i][j]+d[1][i][j]+d[2][i][j]-3+1;
                ans=min(ans,tmp);
            }
        }
        if(ans==INF) puts("-1");
        else cout<<ans<<endl;
    }
    return 0;
}

View Code

做题稳定性有待提升。

posted @ 2015-10-27 04:07 __560 阅读(212) 评论(0) 编辑收藏举报

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codeforces#327 div2

codeforces#327 div2

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