ACM ICPC 2017 Warmup Contest 5

据说似乎是挂错题了，三道水题，估计是某比赛的热身赛吧

结果系统还炸了，什么thin pool溢出，什么测评姬无数据，各种ce

好在最后时刻成功ak了，想想这似乎是我除大一上机50+题ak后时隔两年后的第一次ak，玄学

 A. Advice from Jad

思路：简单的字符串匹配，一开始看到的时候，对于字符串渣渣的我来说有点蒙，然而这题确实有点水，一开始以为是个前缀或模式匹配什么的，后来发现kmp都用不到 ，直接O（mn）强行匹配就好了

#include <iostream>#include <cstdio>#include <algorithm>#include <vector>#include <cstring>#include <map>#include <cmath>#include <string>#include <queue>#include <stack>using namespace std;const int maxn = 110;char ep[10][maxn];char icpcp[10][maxn];int len[10];int main(){    int n,m,tt;    while(scanf("%d",&tt)!=EOF)    {        for(int t=1;t<=tt;t++)        {            scanf("%d%d",&n,&m);            getchar();            for(int i=0;i<n;i++)            {                gets(ep[i]);                len[i] = strlen(ep[i]);            }            for(int i=0;i<m;i++)            {                gets(icpcp[i]);            }            printf("Case #%d:\n",t);            for(int i=0;i<m;i++)            {                bool diff = true;                int icpclen = strlen(icpcp[i]);                //cout << icpclen << "*"<<endl;                for(int j=0;diff&&j<n;j++)                {                    int pos = 0;                    for(int k=0;k<icpclen&&pos<len[j];k++)                    {                        if(icpcp[i][k] != ep[j][pos])                        {                            break;                        }                        else                        {                            pos++;                        }                    }                    if(pos >= len[j])                    {                        diff = false;                    }                    //cout << pos << "***" << endl;                }                if(diff)                {                    printf("This problem may be somewhat difficult.\n");                }                else                {                    printf("This is an easy problem.\n");                }            }            printf("\n");        }    }    return 0;}

C. Chat Parser

思路：其实这题和上题差不多，也是字符串匹配，就是这回按行读，且不知道串长，迫不得已开string，用cin，关同步，并存储于map，就ok了

#include <iostream>#include <cstdio>#include <algorithm>#include <vector>#include <cstring>#include <map>#include <cmath>#include <string>#include <queue>#include <stack>using namespace std;const int maxn = 1e5+10;string s;map<string,bool> p;int main(){ios::sync_with_stdio(false);     int tt;    while(cin>>tt)    {        getchar();        for(int t=1;t<=tt;t++)        {            p.clear();            while(getline(cin,s))            {                int len = s.size();                if(len==0)                {                    break;                }                int pos;                for(pos=0;pos<len;pos++)                {                    if(s[pos] == ':')                        break;                }                string ns = s.substr(0,pos);                p[ns] = true;            }            cout << "Case #" << t << ": " << p.size()<<endl;        }    }    return 0;}

 B. Boats on the lake

思路：计算几何，判断点在多边形内部，利用射线交点奇偶性来判断，对于边缘点，特判点在直线上就好了，感觉计算几何这块可以稍微整理一下，但可能意义不是很大，据说现场赛计算几何的难度都属于不可做，看情况吧

#include <iostream>#include <cstdio>#include <algorithm>#include <vector>#include <cstring>#include <map>#include <cmath>#include <string>#include <queue>#include <stack>using namespace std;const int maxn = 110;double bx[maxn],by[maxn],polyX[maxn],polyY[maxn];bool pointInPolygon(int ployCorners,double x,double y);int main(){    int tt;    while(scanf("%d",&tt)!=EOF)    {        for(int t=1;t<=tt;t++)        {            printf("Case #%d:\n",t);            int n;            scanf("%d",&n);            for(int i=1;i<=n;i++)            {                scanf("%lf%lf",&bx[i],&by[i]);            }            int m;            scanf("%d",&m);            for(int i=0;i<m;i++)            {                int sum = 0;                int p;                scanf("%d",&p);                for(int j=0;j<p;j++)                {                    int pos;                    scanf("%d",&pos);                    polyX[j] = bx[pos];                    polyY[j] = by[pos];                }                for(int j=1;j<=n;j++)                {                    if(pointInPolygon(p,bx[j],by[j]))                    {                        sum++;                    }                }               // cout << pointInPolygon(p,tx,ty,0,15) << endl;                /*for(int j=0;j<p;j++)                {                    cout << tx[j] << " " << endl;                }*/                printf("%d\n",sum);            }        }    }    return 0;}//  Globals which should be set before calling this function:////  int    polyCorners  =  how many corners the polygon has//  float  polyX[]      =  horizontal coordinates of corners//  float  polyY[]      =  vertical coordinates of corners//  float  x, y         =  point to be tested////  (Globals are used in this example for purposes of speed.  Change as//  desired.)////  The function will return YES if the point x,y is inside the polygon, or//  NO if it is not.  If the point is exactly on the edge of the polygon,//  then the function may return YES or NO.////  Note that division by zero is avoided because the division is protected//  by the "if" clause which surrounds it.double getdis(double ax,double ay,double bx,double by){    return sqrt((ax-bx)*(ax-bx) + (ay-by)*(ay-by));}bool inLine(double px,double py,double ax,double ay,double bx,double by){    //cout << ax << "^^" << ay << "^^" << bx << "^^" << by << endl;    double ap = getdis(ax,ay,px,py);    double bp = getdis(bx,by,px,py);    double ab = getdis(ax,ay,bx,by) + 1e-8;    if(ap+bp<=ab)    {        //cout << ap << "#" << bp << "#" << ab << endl;        return true;    }    else    {        //cout << ap << "#" << bp << "#" << ab << endl;        return false;    }}bool pointInPolygon(int polyCorners,double x,double y){    /*for(int j=0;j<polyCorners;j++)            {                cout << polyX[j] << " " << endl;            }*/    int   i, j=polyCorners-1 ;    for(i=0;i<j;i++)    {        if(inLine(x,y,polyX[i],polyY[i],polyX[i+1],polyY[i+1]))        {            //cout << "*"<<i+1 <<"*"<<x<<"*"<<y<<"*"<<polyX[i]<<"*"<<polyY[i]<<"*"<<polyX[i+1]<<"*"<<polyY[i+1]<< endl;            return true;        }    }    if(inLine(x,y,polyX[j],polyY[j],polyX[0],polyY[0]))    {        return true;    }    bool  oddNodes=false ;    for (i=0; i<polyCorners; i++)    {        if (((polyY[i]< y && polyY[j]>=y)        ||   (polyY[j]< y && polyY[i]>=y))        &&   (polyX[i]<=x || polyX[j]<=x) )        {            oddNodes^=(polyX[i]+(y-polyY[i])/(polyY[j]-polyY[i])*(polyX[j]-polyX[i])<x);        }        j=i;    }    //cout << "**" << endl;    return oddNodes;}

文章地址：http://blog.csdn.net/owen_q/article/details/78170302