130831组队赛-Regionals 2011, Asia - Kuala Lumpur

A.Smooth Visualization

简单模拟题，不多说了。

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<string>#include<cmath>#include<set>#include<vector>#include<stack>#define mem(a,b) memset(a,b,sizeof(a))#define FOR(a,b,i) for(i=a;i<=b;++i)#define For(a,b,i) for(i=a;i<b;++i)#define N 1000000007using namespace std;inline void RD(int &ret){    char c;    do    {        c=getchar();    }    while(c<'0'||c>'9');    ret=c-'0';    while((c=getchar())>='0'&&c<='9')    {        ret=ret*10+(c-'0');    }}inline void OT(int a){    if(a>=10)    {        OT(a/10);    }    putchar(a%10+'0');}int n;char s[105];int ss[705];int main(){    int t;    scanf("%d",&t);    getchar();    while(t--)    {        gets(s);        int l = strlen(s);        int ll = 0;        int ma = 0;        int a = s[0] - '0';        ss[0] = a;        ma = a;        ll++;        int b;        for(int i=1;i<l;i++)        {            a = s[i-1] - '0';            b = s[i] - '0';            ma = max(ma,b);            if(b == a)            {                ss[ll] = b;                ll++;            }            if(b > a)            {                int c = a + 1;                do                {                    ss[ll] = c;                    c++;                    ll++;                }while(c <= b);            }            if(b < a)            {                int c = a - 1;                do                {                    ss[ll] = c;                    c--;                    ll++;                }while(c >= b);            }        }        for(int i = ma;i >=1;i--)        {            for(int j=0;j<ll;j++)            {                if(i > ss[j]) printf("*");                else printf("+");            }            printf("\n");        }    }    return 0;}

E.Social Holidaying

二分图匹配，题意是给你两个序列，要求A序列中最多有多少对数和为B序列中的数，且A序列中的数每个只能用一次，B序列的数可以无限次出现。

思路：可以将A序列中所有能组成的可能全都组合起来，如果B序列中有出现就开始搜索，把所有情况都搜出来。

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<string>#include<cmath>#include<set>#include<vector>#include<stack>#include<map>#define mem(a,b) memset(a,b,sizeof(a))#define FOR(a,b,i) for(i=a;i<=b;++i)#define For(a,b,i) for(i=a;i<b;++i)#define N 1000000007using namespace std;inline void RD(int &ret){    char c;    do    {        c=getchar();    }    while(c<'0'||c>'9');    ret=c-'0';    while((c=getchar())>='0'&&c<='9')    {        ret=ret*10+(c-'0');    }}inline void OT(int a){    if(a>=10)    {        OT(a/10);    }    putchar(a%10+'0');}map<long long,int>s;vector<int>p[1001];int vis[1001],tmp[1001];long long x[1001],y;int f(int j){    int i,q;    for(i=0;i<p[j].size();++i)    {        q=p[j][i];        if(vis[q]==0)        {            vis[q]=1;            if(tmp[q]==0||f(tmp[q])==1)            {                tmp[q]=j;                return 1;            }        }    }    return 0;}int main(){    int t,i,j,sum,n,m;    scanf("%d",&t);    while(t--)    {        s.clear();        scanf("%d%d",&n,&m);        for(i=0;i<n;++i)        {            scanf("%lld",&x[i]);        }        for(i=0;i<m;++i)        {            scanf("%lld",&y);            s[y]=1;        }        for(i=0;i<n;++i)        {            p[i].clear();        }        sum=0;        for(i=0;i<n;++i)        {            for(j=0;j<n;++j)            {                if(i==j)                {                    continue;                }                if(s[x[i]+x[j]]==1)                {                    p[i].push_back(j);                }            }        }        mem(tmp,0);        for(i=0;i<n;++i)        {            mem(vis,0);            if(f(i)==1)            {                sum++;            }        }        printf("%d\n",sum/2);    }    return 0;}

G.Writings on the Wall

字符串问题，问的是前后两个字符串最多能重合部分几次。

方法有两种：kmp和hash

KMP写法：

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<string>#include<cmath>#include<set>#include<vector>#include<stack>#define mem(a,b) memset(a,b,sizeof(a))#define FOR(a,b,i) for(i=a;i<=b;++i)#define For(a,b,i) for(i=a;i<b;++i)//#define N 1000000007using namespace std;inline void RD(int &ret){    char c;    do    {        c=getchar();    }    while(c<'0'||c>'9');    ret=c-'0';    while((c=getchar())>='0'&&c<='9')    {        ret=ret*10+(c-'0');    }}inline void OT(int a){    if(a>=10)    {        OT(a/10);    }    putchar(a%10+'0');}#define N 50005char a[N];char b[N];int next[N];int l1,l2;void update(){    next[0] = -1;    int i=0,j=-1;    while(i<l2)    {        while(j>=0&&b[i]!=b[j])        {            j=next[j];        }        i++;        j++;        next[i]=j;    }}int main(){    int t,i,j,ans;    scanf("%d",&t);    while(t--)    {        scanf("%s%s",a,b);        l1=strlen(a);        l2=strlen(b);        update();        i=0;        j=0;        while(i<l1)        {            while(j>=0&&a[i]!=b[j])            {                j=next[j];            }            i++;            j++;            if(j==l2&&i<l1)            {                j=next[j];            }        }        ans=0;        while(j>=0)        {            j=next[j];            ans++;        }        printf("%d\n",ans);    }    return 0;}

HASH写法：

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<string>#include<cmath>#include<set>#include<vector>#include<stack>#include <queue>#include<map>#define mem(a,b) memset(a,b,sizeof(a))#define FOR(a,b,i) for(i=a;i<=b;++i)#define For(a,b,i) for(i=a;i<b;++i)using namespace std;inline void RD(int &ret){    char c;    do    {        c=getchar();    }    while(c<'0'||c>'9');    ret=c-'0';    while((c=getchar())>='0'&&c<='9')    {        ret=ret*10+(c-'0');    }}inline void OT(int a){    if(a>=10)    {        OT(a/10);    }    putchar(a%10+'0');}int main(){    int i,j,l1,l2,t,sum;    long long x,y,z;    char a[50001],b[50001];    RD(t);    while(t--)    {        scanf("%s%s",a,b);        l1=strlen(a);        l2=strlen(b);        i=l1-1;        j=0;        z=1;        x=0;        y=0;        sum=1;        while(i>=0&&j<l2)//hash过程        {            x+=z*(a[i]-'a');            z*=33;            y*=33;            y+=(b[j]-'a');            i--;            j++;            if(x==y)            {                sum++;            }        }        printf("%d\n",sum);    }    return 0;}

H.Robotic Traceur

最短路，我们悲剧地居然一直TLE，后来才发现BFS，SPFA，甚至连暴搞都可以过。。。

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<string>#include<cmath>#include<set>#include<vector>#include<stack>#include <queue>#include<map>#define mem(a,b) memset(a,b,sizeof(a))#define FOR(a,b,i) for(i=a;i<=b;++i)#define For(a,b,i) for(i=a;i<b;++i)using namespace std;inline void RD(int &ret){    char c;    do    {        c=getchar();    }    while(c<'0'||c>'9');    ret=c-'0';    while((c=getchar())>='0'&&c<='9')    {        ret=ret*10+(c-'0');    }}inline void OT(int a){    if(a>=10)    {        OT(a/10);    }    putchar(a%10+'0');}struct xl{    double x,y;}v[1001];double dis(int i, int j){    return sqrt((v[i].x-v[j].x)*(v[i].x-v[j].x)+(v[i].y-v[j].y)*(v[i].y-v[j].y));}int main(){    int t,n,s,tmp,i,id[1001],l[1001],w,res;    double p,q;    RD(t);    while(t--)    {        scanf("%d%d%d%lf%lf",&n,&s,&tmp,&p,&q);        p+=q;        FOR(1,n,i)        {            scanf("%lf%lf",&v[i].x,&v[i].y);            id[i]=-1;        }        id[s]=0;        l[0]=s;        w=1;        res=0;        while(res<w)        {            s=l[res];            FOR(1,n,i)            {                if(id[i]<0&&dis(s,i)<p+1e-6)                {                    id[i]=id[s]+1;                    l[w++]=i;                    if(i==tmp)                    {                        break;                    }                }            }            res++;        }        if(id[tmp]>=0)        {            printf("%d\n",id[tmp]);        }        else        {            printf("Impossible\n");        }    }    return 0;}

I.Shortest Leash

应该是一道dp题，它要求狗可以选择多个向量的正负方向走，问最远能与远点的距离为多少。。。

我是用随机算法过的，目的在于将所有的向量打乱，使其在加上向量时，能够会在一次中找到正解。

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<string>#include<cmath>#include<set>#include<vector>#include<stack>#include<map>#define mem(a,b) memset(a,b,sizeof(a))#define FOR(a,b,i) for(i=a;i<=b;++i)#define For(a,b,i) for(i=a;i<b;++i)using namespace std;struct xl{    double x,y;} p[101],q,l,r;double dis(xl a){    return sqrt(a.x*a.x+a.y*a.y);}int main(){    int t,i,j,z[101];    double ans;    while(scanf("%d",&t))    {        if(t==0)        {            break;        }        for(i=0; i<t; ++i)        {            scanf("%lf%lf",&p[i].x,&p[i].y);        }        ans=0;        for(j=0;j<=10000;++j)        {            random_shuffle(p,p+t);//随机打乱数组            q.x=0;            q.y=0;            for(i=0; i<t; ++i)            {                l=q;                r=q;                l.x+=p[i].x;                l.y+=p[i].y;                r.x-=p[i].x;                r.y-=p[i].y;                if(dis(l)>dis(r))                {                    q=l;                }                else                {                    q=r;                }                ans=max(ans,dis(q));            }        }        printf("%.3f\n",ans);    }    return 0;}