ACM ICPC 2017 Warmup Contest 7（CTU Open Contest 2016）

ccsp与区域赛都越来越近了，模拟与区域赛题并进，还有一堆作业，有点累，想玩耍，感觉自己有点迷失，算了，还是就这样吧，努力向前

练习赛7，打两个签到题走人，继续刷csp去

B. Hot Air Ballooning

思路：统计不同人用过的气球的方案数，又是个去重问题，又想往set上放，后来发现气球数很少，完全可以数组统计，而气球总组合有限，虽然不大，但强搜可能会感觉tle，加个状压好了，感觉现在自己特别喜欢做状压的题，特别得心应手

#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 = 1010;bool situation[maxn];bool temp[10];char temps[maxn];int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        int sum = 0;        memset(situation,false,sizeof(situation));        while(n--)        {            scanf("%s",temps);            memset(temp,false,sizeof(temp));            int len = strlen(temps);            for(int i=0;i<len;i++)            {                    temp[temps[i]-'0'] = true;            }            int now = 0;            for(int pos=1;pos<=9;pos++)            {                if(temp[pos])                {                    now |= (1<<(pos-1));                }            }            //cout << now << " * " << endl;            if(situation[now]==false)            {                situation[now] = true;                sum++;            }        }        printf("%d\n",sum);    }    return 0;}

J. Colorful Tribune

思路：不过这次的签到题似乎有点过于水了，n行n列n类数每类n个，每行每列不重，有一个特殊元素，找出并复原

这要分两种情况，一个是特殊元素为新加入的元素，二是特殊元素为已有元素，第二种情况相对比较负责，需要重新定位，最后还是一点小细节吧

#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 = 100;char tribune[maxn][maxn];int num[maxn];int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        bool one = false;        memset(num,0,sizeof(num));        for(int i=0;i<n;i++)        {            scanf("%s",tribune[i]);            for(int j=0;j<n;j++)            {                num[tribune[i][j]-'A']++;            }        }        char wrong,correct;        for(int i=0;i<26;i++)        {            if(num[i]==1)            {                wrong = i + 'A';                one = true;            }            else if(num[i] == n+1)            {                wrong = i + 'A';            }            else if(num[i] == n-1)            {                correct = i + 'A';            }        }        //cout << wrong << endl;        if(one)        {            for(int i=0;i<n;i++)            {                for(int j=0;j<n;j++)                {                    if(tribune[i][j]==wrong)                    {                        printf("%d %d %c\n",i+1,j+1,correct);                    }                }            }        }        else        {            int line[maxn];            int row[maxn];            memset(line,0,sizeof(line));            memset(row,0,sizeof(row));            for(int i=0;i<n;i++)            {                for(int j=0;j<n;j++)                {                    if(tribune[i][j]==wrong)                    {                        line[i]++;                        row[j]++;                    }                }            }            int rel,rer;            for(int i=0;i<n;i++)            {                if(line[i]>1)                {                    rel = i+1;                }                if(row[i]>1)                {                    rer = i+1;                }                //cout << line[i] << "*" << row[i]<< endl;            }            printf("%d %d %c\n",rel,rer,correct);        }    }    return 0;}/*6OEYCDKEYOKCDKDCEOYCKHOYEYOEDKCDCKYEO3IWLGIWWLI6OEYCDKEYOKCDKDCEOYCKKOYEYOEDKCDCKYEO*/

好了，继续回去刷csp了

补完一道csp的T3大模拟后，还是觉得补一道icpc的题换换吧

 I. Suspicious Samples

思路：其实这题的难度不大，但通过这题，对于set的插入、删除和排序又有了新的认识，确实收获不少

set中cmp的编写

这个确实是一个非常关键的地方，第一是书写格式，第二就是一定要确定对于所有满足的情况全部都true；

由于set去、具有去重性，所以一点点小小的疏忽就会导致部分值部莫名其妙地去掉了

struct cmp{    bool operator()(Pattern a,Pattern b)const    {        if((a.value>b.value)||(a.value==b.value&&a.time<b.time))        {            return true;        }        else        {            return false;        }    }};

还有就是earse了，这是一个大坑点啊

1.earse完后会导致迭代器失效，所以一定要注意及时重新对迭代器赋值（这题没遇到这个坑点，但确实是常见而又易错的坑点）

2.earse和find结合时，要注意find返回值为末尾时（为找到），千万要特判，不可随意earse

3.一旦set中的元素被earse完后，所有begin（），end（）及类似迭代器全部失效，这个算是今天最大的坑点了吧

if(pattern.find(tempdelete)!=pattern.end()){       pattern.erase(pattern.find(tempdelete));}if(pattern.empty()){       now = 0;}else{       ……}

还有就是界限退出条件的判断这些了吧

#include <iostream>#include <cstdio>#include <algorithm>#include <vector>#include <cstring>#include <map>#include <cmath>#include <string>#include <queue>#include <stack>#include <set>using namespace std;const int maxn = 1e5+10;int t[maxn];int v[maxn];typedef struct PATTERN{    int time;    int value;}Pattern;struct cmp{    bool operator()(Pattern a,Pattern b)const    {        if((a.value>b.value)||(a.value==b.value&&a.time<b.time))        {            return true;        }        else        {            return false;        }    }};set<Pattern,cmp> pattern;int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        t[0] = -1e9;        v[0] = 0;        for(int i=1;i<=n;i++)        {            scanf("%d%d",&t[i],&v[i]);        }        int c;        scanf("%d",&c);        while(c--)        {            pattern.clear();            int re = 0;            char r[10],f[10];            int l;            scanf("%s%s%d",r,f,&l);            int pos,st=1;            int sum = 0;            int num = 0;            for(pos=2;pos<=n;pos++)            {                //cout << pos << "*" <<st<< endl;                sum += v[pos-1];                num++;                //cout << pos << "#" << st << endl;                //set<Pattern> :: iterator it;                Pattern temp;                temp.value = v[pos-1];                temp.time = t[pos-1];                pattern.insert(temp);                /*cout << pos << "#" << st <<"#" << t[pos-1]<< endl;                for(it=pattern.begin();it!=pattern.end();it++)                {                    cout << (it->time) << " ";                }                cout <<endl;*/                while(t[pos]-t[st]>l&&st<pos)                {                    //cout << pos << "#" << st << endl;                    sum -= v[st];                    num--;                    Pattern tempdelete;                    tempdelete.value = v[st];                    tempdelete.time = t[st];                    //cout << "^" <<pattern.size()<<"^"<<((pattern.find(tempdelete))->time)<< endl;                    if(pattern.find(tempdelete)!=pattern.end())                    {                        /*if(pattern.size()==1)                        {                            pattern.clear();                        }                        else                        {*/                            pattern.erase(pattern.find(tempdelete));                        //}                    }                    //cout << "^^" << endl;                    /*cout << pos << "^" << st <<"^" << t[st]<< endl;                    for(it=pattern.begin();it!=pattern.end();it++)                    {                        cout << (it->time) << " ";                    }                    cout <<endl;*/                    st++;                }                //cout << "&&" << endl;                long long now;                if(pattern.empty())                {                    now = 0;                }                else                {                    if(strcmp(f,"max")==0)                    {                        now = (long long)((pattern.begin())->value) * (long long)(num);                    }                    else if(strcmp(f,"min")==0)                    {                        now = (long long)((pattern.rbegin())->value) * (long long)(num);                    }                    else                    {                        now = (long long)(sum);                    }                }                if(strcmp(r,"gt")==0&&((long long)(v[pos])*(long long)(num)>now))                {                    re++;                }                else if(strcmp(r,"lt")==0&&((long long)(v[pos])*(long long)(num)<now))                {                    re++;                }            }            printf("%d\n",re);        }    }    return 0;}

文章地址：http://blog.csdn.net/owen_q/article/details/78234317