Codeforces Round #433 (Div. 2, based on Olympiad of Metropolises)

近期最后一场cf了，双休日icpc网络赛之后，就要开始准备数模国赛了，加油吧

A. Fraction

思路：寻找最大最简真分数，题有点长，不过理解后难度不大，枚举判断gcd即可

#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;int gcd(int a,int b){return b == 0 ? a : gcd(b, a % b);}int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        for(int a = n/2;a>=1;a--)        {            int b = n - a;            if(gcd(a,b)==1)            {                printf("%d %d\n",a,b);                break;            }        }    }    return 0;}

B. Maxim Buys an Apartment

思路：寻找最优房，样例有很大提示性作用，所有空房连一起肯定是最差策略，一个租房连两个空房肯定是最优策略，最后特判一下无房情况即可

#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;int main(){    int n,k;    while(scanf("%d%d",&n,&k)!=EOF)    {        if(k==0||k==n)        {            printf("0 0\n");        }        else        {            int mininum = 1;            int maxinum = 0;            if(2*k>=n-k)            {                maxinum = n-k;            }            else            {                maxinum = 2 * k;            }            printf("%d %d\n",mininum,maxinum);        }    }    return 0;}

C. Planning

思路：这题就比较坑了，如果想明白就很简单

简单的并查集问题，每个时间点作为一个集合，将无法起飞的集合并入下一个集合

贪心按延期代价排序，最先满足代价高的，该时间段所属集合即为起飞时间，每次log的复杂度

纯并查集裸题，然而，思路一般不容易想到，大量的时间细化杂乱的细节，最后却发现并查集完美解决

#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 = 3e5+10;typedef struct FLIGHT{    int time;    int c;}Flight;Flight f[maxn];int pre[2*maxn];int takeoff[maxn];//bool takeoff[2*maxn];int finds(int x){ //寻找    int r = x;    while(pre[r]!=r){        r = pre[r];    }    int i=x ,j;    while(pre[i]!=r){ //路径压缩        j = pre[i];        pre[i] = r;        i = j;    }    return r;}void join(int x,int y){ //将两个集合合并    int fx = finds(x);    int fy = finds(y);    if(fx!=fy){        pre[fx] = fy;    }}bool cmp(const Flight &a,const Flight &b){    if(a.c>b.c)    {        return true;    }    else    {        return false;    }}int main(){    int n,k;    while(scanf("%d%d",&n,&k)!=EOF)    {        //memset(takeoff,false,sizeof(takeoff));        //takeoff[k+1] = true;        pre[n+k+1] = n+k+1;        for(int i=1;i<=n;i++)        {            scanf("%d",&f[i].c);            f[i].time = i;            pre[i] = i;            /*else            {                takeoff[i+1] = true;            }*/        }        for(int i=n+1;i<=k+n;i++)        {            pre[i] = i;            //takeoff[i] = true;        }        for(int i=1;i<=k;i++)        {            join(i,k+1);        }        sort(f+1,f+n+1,cmp);        long long sum = 0;        for(int i=1;i<=n;i++)        {            //cout << f[i].c << endl;            /*if(takeoff[f[i].time])            {                takeoff[f[i].time] = false;                join(f[i].time,f[i].time+1);            }            else            {            }*/            sum += (long long)(finds(f[i].time) - f[i].time) * (long long)(f[i].c);            /*cout << sum << endl;            for(int j=1;j<=8;j++)            {                cout << pre[j] << " ";            }            cout << endl;            cout << f[i].time<<"*"<<finds(f[i].time)+1<<endl;*/            takeoff[f[i].time] = finds(f[i].time);            join(f[i].time,finds(f[i].time)+1);        }        printf("%I64d\n",sum);        for(int i=1;i<n;i++)        {            printf("%d ",takeoff[i]);        }        printf("%d\n",takeoff[n]);    }    return 0;}

最后，比赛加油

文章地址：http://blog.csdn.net/owen_q/article/details/77878567