上海邀请赛

2014-上海邀请赛

A Game with Pearls 

传送门:hdu 5090

给定一个长度为n的数组,可以对任意一个数增加k的倍数,问能否通过调换位置,增数来使得整个数组构成一个1～n的序列

贪心，对k取模分组,因为加k取模不会变,所以首先判断对应组内的个数是否满足最终序列需要的个数,然后从小到大分别对各组内进行判断

/****************************************************** * File Name:   1001.cpp * Author:      kojimai * Create Time: 2014年11月02日 星期日 12时47分52秒******************************************************/#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<iostream>#include<vector>using namespace std;#define FFF 105vector<int> p[FFF];int a[FFF];int main(){int keng;scanf("%d",&keng);while(keng--){int n,k;cin>>n>>k;for(int i = 0;i < n;i++)cin>>a[i];sort(a,a+n);for(int i = 0;i < k;i++){p[i].clear();}for(int i = 0;i < n;i++){int t = a[i]%k;p[t].push_back(a[i]);}bool flag = true;for(int i = 0;i < k && flag;i++){int cnt = n / k;if(i && i <= n % k)cnt++;if(cnt != p[i].size())flag= false;}if(flag)for(int i = 1;i < n&&flag;i++){int t = i % k;if(p[t][0]<=i){p[t].erase(p[t].begin());}elseflag = false;}if(flag)printf("Jerry\n");elseprintf("Tom\n");}return 0;}

C Seam Carving 

传送门:hdu 5092

题意比较绕,理解了题意就好做了,从上到下找一条路径,使得路径上的值最小,要求相邻行之间选到的点左右距离值不大大于1

dp类似数塔,求最小统计路径即可

P.S.上海我们队最后一小时就卡在上面了,理解了题意就是敲不出来,做重现顺利过,简直无情.....

/****************************************************** * File Name:   1003.cpp * Author:      kojimai * Create Time: 2014年11月02日 星期日 14时25分57秒******************************************************/#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<iostream>using namespace std;#define FFF 105long long a[FFF][FFF],dp[FFF][FFF];int road[FFF][FFF];int main(){int keng,Case = 1;cin>>keng;while(keng--){int n,m;cin>>n>>m;for(int i = 1;i <= n;i++){for(int j = 1;j <= m;j++){cin>>a[i][j];}}memset(dp,0,sizeof(dp));for(int i = 1;i <= m;i++){dp[n][i] = a[n][i];road[n][i] = -1;}for(int i = n - 1;i >= 1;i--){for(int j = m;j >= 1;j--){if(j < m){dp[i][j] = (dp[i+1][j+1] + a[i][j]);road[i][j] = j+1;if(dp[i+1][j]<dp[i+1][j+1]){road[i][j] = j;dp[i][j] = dp[i+1][j]+a[i][j];}}else{dp[i][j] = dp[i+1][j] +a[i][j];road[i][j] = j;}if(j>1){if(dp[i][j]>(dp[i+1][j-1]+a[i][j])){road[i][j] = j-1;dp[i][j] = dp[i+1][j-1]+a[i][j];}}}}long long ans = dp[1][m];int s = m;for(int i = m - 1;i >= 1;i--){if(dp[1][i] < ans){ans = dp[1][i];s = i;}}printf("Case %d\n%d",Case++,s);int cnt = 1;while(road[cnt][s] != -1){s = road[cnt++][s];cout<<' '<<s;}cout<<endl;}return 0;}

D Battle ships 

传送门:hdu 5093

给定一张图,在*位置可以放点,O表示不可放点但点之间可以通过该点相连,#表示可以隔断点问最多放多少个点,使得两两之间不可达

二分图匹配,对每个点进行拆点,分行与列对点编号,同一行或同一列中,中间有#则在行或者列上,该点拆成多个点,这样拆完点之后得到一张二分图进行一次二分匹配即可

/****************************************************** * File Name:   1004.cpp * Author:      kojimai * Create Time: 2014年11月02日 星期日 13时12分13秒******************************************************/#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<iostream>using namespace std;#define FFF 55char s[FFF][FFF];int hang[FFF][FFF],lie[FFF][FFF];#define FF 1300int match[FF],cntx,cnty;bool mm[FF][FF],vis[FF];bool dfs(int now){for(int i = 0;i < cnty;i++){if(mm[now][i]&&!vis[i]){vis[i] = true;if(match[i] == -1 || dfs(match[i])){match[i] = now;return true;}}}return false;}void printhang(){for(int i = 0;i < 4;i++){for(int j = 0;j < 4;j++)cout<<hang[i][j]<<' ';cout<<endl;}}void printlie(){for(int i = 0;i < 4;i++){for(int j = 0;j < 4;j++)cout<<lie[i][j]<<' ';cout<<endl;}}int main(){int keng;cin>>keng;while(keng--){int n,m;scanf("%d%d",&n,&m);for(int i = 0;i < n;i++)cin>>s[i];cntx = cnty = 0;bool flag = false;memset(hang,-1,sizeof(hang));memset(lie,-1,sizeof(lie));for(int i = 0;i < n;i++){for(int j = 0;j < m;j++){if(s[i][j] == '*'){hang[i][j] = cntx;flag = true;}else if(s[i][j] == '#'){if(flag == true){flag = false;cntx++;}}}if(flag){flag = false;cntx++;}}//在行上拆点//printhang();flag = false;for(int i = 0;i < m;i++){for(int j = 0;j < n;j++){if(s[j][i] == '*'){lie[j][i] = cnty;flag = true;}else if(s[j][i] == '#'){if(flag){flag = false;cnty++;}}}if(flag){flag = false;cnty++;}}//在列上拆点//printlie();//cout<<" cntx = "<<cntx<<" cnty = "<<cnty<<endl;memset(mm,false,sizeof(mm));for(int i = 0;i < n;i++){for(int j = 0;j < m;j++){if(s[i][j] == '*'){mm[hang[i][j]][lie[i][j]] = true;}}}//for(int i = 0;i < cntx;i++)//{//for(int j = 0;j < cnty;j++)//cout<<mm[i][j]<<' ';//cout<<endl;//}memset(match,-1,sizeof(match));int ans = 0;for(int i = 0;i < cntx;i++){memset(vis,false,sizeof(vis));if(dfs(i))ans++;}//答案即为最大匹配数cout<<ans<<endl;}return 0;}

F  Linearization of the kernel functions in SVM 

传送门:hdu 5095

签到题,注意各种边界条件处理即可

/****************************************************** * File Name:   1006.cpp * Author:      kojimai * Create Time: 2014年11月02日 星期日 13时39分25秒******************************************************/#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<iostream>using namespace std;bool first;void printnum(int x){if(first){first = false;if(x>0){if(x == 1)return;elsecout<<x;}else{if(x == -1)cout<<'-';elsecout<<x;}}else{if(x>0){if(x == 1)cout<<'+';elsecout<<'+'<<x;}else{if(x == -1)cout<<'-';elsecout<<x;}}return;}void printp(int x){if(x == 0)return;printnum(x);cout<<'p';}void printq(int x){if(x == 0)return;printnum(x);cout<<'q';}void printr(int x){if(x == 0)return;printnum(x);cout<<'r';}void printu(int x){if(x == 0)return;printnum(x);cout<<'u';}void printv(int x){if(x == 0)return;printnum(x);cout<<'v';}void printw(int x){if(x == 0)return;printnum(x);cout<<'w';}void printx(int x){if(x == 0)return;printnum(x);cout<<'x';}void printy(int x){if(x == 0)return;printnum(x);cout<<'y';}void printz(int x){if(x == 0)return;printnum(x);cout<<'z';}int main(){int keng;cin>>keng;while(keng--){int p,q,r,u,v,w,x,y,z,j;cin>>p>>q>>r>>u>>v>>w>>x>>y>>z>>j;first = true;printp(p);printq(q);printr(r);printu(u);printv(v);printw(w);printx(x);printy(y);printz(z);if(j){if(!first&&j>0)cout<<'+'<<j;elsecout<<j;first = false;}if(first)cout<<0<<endl;cout<<endl;}return 0;}

J Comparison of Android versions

传送门:hdu 5099

水题,比较大小就好,一个注意点,分支相同时,在时间比较上才需要考虑最后一个字母

/****************************************************** * File Name:   1010.cpp * Author:      kojimai * Create Time: 2014年11月02日 星期日 14时09分47秒******************************************************/#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<iostream>using namespace std;char s1[10],s2[10];int main(){int keng,Case =1;cin>>keng;while(keng--){int num1,num2;cin>>s1>>s2;if(s1[0] == s2[0]){num1 =0 ;}else if(s1[0] > s2[0])num1 = 1;elsenum1 = -1;if(s1[2] < s2[2])num2 = -1;else if(s1[2] > s2[2])num2 = 1;else{int tmp1 = (s1[3]-'0')*10+s1[4]-'0';int tmp2 = (s2[3]-'0')*10+s2[4]-'0';if(tmp1>tmp2)num2 = 1;else if(tmp1<tmp2)num2 = -1;else {num2 = 0;}}if(num2 == 0 && s1[1] == s2[1]){if(s1[5]>s2[5]){num2 = 1;}else if(s1[5]<s2[5])num2 = -1;}printf("Case %d: ",Case++);if(num1==1)cout<<'>';else if(num1 == 0)cout<<'=';elsecout<<'<';cout<<' ';if(num2==1)cout<<'>';else if(num2==-1)cout<<'<';elsecout<<'=';cout<<endl;}return 0;}