A

 #include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
using namespace std;
#define N 100000
#define LL long long
#define INF 0xfffffff
const double eps = 1e-;
const double pi = acos(-1.0);
const double inf = ~0u>>;
int main()
{
int n,i,j;
int ans = ;
cin>>n;
for(i = ; i <= n; i++)
{
int x,y;
scanf("%d%d",&x,&y);
if(y-x>=) ans++;
}
cout<<ans<<endl;
return ;
}

B

还以为是间接的朋友,写复杂了,没想到那么简单。

 #include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
using namespace std;
#define N 1010
#define LL long long
#define INF 0xfffffff
const double eps = 1e-;
const double pi = acos(-1.0);
const double inf = ~0u>>;
int a[N];
int fa[N];
int r[N];
int find(int x)
{
if(x!=fa[x])
{
fa[x] = find(fa[x]);
return fa[x];
}
return x;
}
int main()
{
int n,m,k,i,j;
cin>>n>>m>>k;
for(i = ;i <= m+; i++) {fa[i] = i;r[i] = ;}
for(i = ; i <= m+; i++)
scanf("%d",&a[i]);
int ans = ;
for(i = ; i <= m; i++)
{
int cnt = ;
for(int g = ; g < n ;g++)
if((a[m+]&(<<g))!=(a[i]&(<<g))) cnt++;
if(cnt<=k) ans++;
}
// for(i = 1; i <= m+1; i++)
// for(j = 1; j <= m+1; j++)
// {
// if(i==j) continue;
// int cnt = 0;
// for(int g = 0 ; g < n ;g++)
// if((i&(1<<g))!=(i&(1<<g))) cnt++;
// if(cnt<=k)
// {
// int tx = find(i);
// int ty = find(j);
// if(tx!=ty)
// {
// fa[tx] = ty;
// r[tx]+=r[ty];
// }
// }
// }
// int kk = find(m+1);
cout<<ans<<endl;
return ;
}

 C

简单dp

 #include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
using namespace std;
#define N 5010
#define LL long long
#define INF 0xfffffff
const double eps = 1e-;
const double pi = acos(-1.0);
const double inf = ~0u>>;
int a[N];
LL dp[N][N];
LL sum[N];
int main()
{
int n,m,k,i,j;
cin>>n>>m>>k;
for(i = ; i <= n; i++)
{
scanf("%d",&a[i]);
sum[i] = sum[i-]+a[i];
}
for(i = m; i <= n ;i++)
{
for(j = ; j <= k; j++)
dp[i][j] = max(dp[i-m][j-]+sum[i]-sum[i-m],dp[i-][j]);
}
cout<<dp[n][k]<<endl;
return ;
}

D

tarjan缩点+dfs

dfs的时候少写了else里面的内容。。一直wa到结束

 #include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#include<stack>
using namespace std;
#define N 500010
#define M 500010
#define LL long long
#define INF 0xfffffff
const double eps = 1e-;
const double pi = acos(-1.0);
const double inf = ~0u>>;
struct node
{
int u,v,next,w;
} edge[M];
int t,low[N],pre[N],sccno[N],head[N],scc,dep,vis[N],dis[N];
int dd1[N],dd2[N];
int dp[N];
int num[N],de[N];
vector<int>cd[N];
void init()
{
t = ;
memset(head,-,sizeof(head));
}
void add(int u,int v)
{
edge[t].u =u;
edge[t].v = v;
edge[t].next = head[u];
head[u] = t++;
}
stack<int>s;
void dfs(int u)
{
low[u] = pre[u] = ++dep;
s.push(u);
for(int i = head[u] ; i!=- ; i = edge[i].next)
{
int v = edge[i].v;
if(!pre[v])
{
dfs(v);
low[u] = min(low[u],low[v]);
}
else if(!sccno[v])
low[u] = min(low[u],pre[v]);
}
if(low[u]==pre[u])
{
scc++;
int minz = INF,sum=INF;
for(;;)
{
int x = s.top();
s.pop();
if(minz>num[x])
{
minz = num[x];
sum = dp[x];
}
else if(minz==num[x]) sum = min(dp[x],sum);
sccno[x] = scc;
if(x==u)break;
}
dd1[scc] = minz;
dd2[scc] = sum; }
}
void find_scc(int n)
{
scc=;
dep=;
memset(low,,sizeof(low));
memset(pre,,sizeof(pre));
memset(sccno,,sizeof(sccno));
for(int i = ; i <= n ; i++)
if(!pre[i])
dfs(i);
}
map<string,int>f;
vector<int>ed[N];
char s1[N],s2[N];
int a[N];
int judge(char *str)
{
int i,k;
k= strlen(str);
int cnt = ;
for(i = ; i < k; i++)
if(str[i]=='R')
cnt++;
return cnt;
}
int ddfs(int u)
{
int i,j;
for(i = ; i< ed[u].size() ; i++)
{
int v = ed[u][i];
if(!vis[v])
{
vis[v] = ;
int ss = ddfs(v);
if(dd1[u]>=ss)
{
if(dd1[u]==ss)
dd2[u] = min(dd2[u],dd2[v]);
else dd2[u] = dd2[v];
dd1[u] = ss;
}
}
else if(dd1[u]>=dd1[v])
{
if(dd1[u]==dd1[v]) dd2[u] = min(dd2[u],dd2[v]);
else dd2[u] = dd2[v];
dd1[u] = dd1[v];
}
}
return dd1[u];
}
int main()
{
init();
int m,i,j,n;
scanf("%d",&m);
int g = ;
for(i = ; i <= m ; i++)
{
scanf("%s",s1);
int len = strlen(s1);
for(j = ; j < len; j++) if(s1[j]>='a'&&s1[j]<='z') s1[j]-=;
if(!f[s1])
{
a[i] = ++g;
f[s1] = g;
}
else a[i] = f[s1];
num[a[i]] = judge(s1);
dp[a[i]] = len;
}
scanf("%d",&n);
for( i = ; i <= n ; i++)
{
int u,v;
scanf("%s%s",s1,s2);
int len1 =strlen(s1) ,len2 = strlen(s2);
for(j = ; j < len1; j++) if(s1[j]>='a'&&s1[j]<='z') s1[j]-=;
for(j = ; j < len2; j++) if(s2[j]>='a'&&s2[j]<='z') s2[j]-=;
if(!f[s1])
{
u = ++g;
f[s1] = g;
}
else u = f[s1];
if(!f[s2])
{
v = ++g;
f[s2] = g;
}
else v = f[s2];
add(u,v);
num[u] = judge(s1);
num[v] = judge(s2);
dp[u] = len1;
dp[v] = len2;
cd[u].push_back(v);
}
find_scc(g);
for(i = ; i <= g; i++)
{
int u = sccno[i];
for(j = ;j < cd[i].size() ; j++)
{
int v = sccno[cd[i][j]];
if(v==u) continue;
ed[u].push_back(v);
de[v] = ;
}
}
memset(vis,,sizeof(vis));
for(i = ; i <= scc; i++)
{
if(!de[i])
{
vis[i] = ;
ddfs(i);
}
}
LL ans = ,sum=;
for(i = ; i<= m; i++)
{
int u = sccno[a[i]];
//cout<<u<<" "<<dd2[u]<<endl;
ans+=dd1[u];
sum+=dd2[u];
}
cout<<ans<<" "<<sum<<endl;
return ;
}

Codeforces Round #267 (Div. 2)的更多相关文章

  1. 01背包 Codeforces Round #267 (Div. 2) C. George and Job

    题目传送门 /* 题意:选择k个m长的区间,使得总和最大 01背包:dp[i][j] 表示在i的位置选或不选[i-m+1, i]这个区间,当它是第j个区间. 01背包思想,状态转移方程:dp[i][j ...

  2. Codeforces Round #267 (Div. 2) C. George and Job(DP)补题

    Codeforces Round #267 (Div. 2) C. George and Job题目链接请点击~ The new ITone 6 has been released recently ...

  3. Codeforces Round #267 Div.2 D Fedor and Essay -- 强连通 DFS

    题意:给一篇文章,再给一些单词替换关系a b,表示单词a可被b替换,可多次替换,问最后把这篇文章替换后(或不替换)能达到的最小的'r'的个数是多少,如果'r'的个数相等,那么尽量是文章最短. 解法:易 ...

  4. Codeforces Round #267 (Div. 2) C. George and Job DP

                                                  C. George and Job   The new ITone 6 has been released ...

  5. Codeforces Round #267 (Div. 2) A

    题目: A. George and Accommodation time limit per test 1 second memory limit per test 256 megabytes inp ...

  6. Codeforces Round #267 (Div. 2)D(DFS+单词hash+简单DP)

    D. Fedor and Essay time limit per test 2 seconds memory limit per test 256 megabytes input standard ...

  7. Codeforces Round #267 (Div. 2) D. Fedor and Essay tarjan缩点

    D. Fedor and Essay time limit per test 2 seconds memory limit per test 256 megabytes input standard ...

  8. Codeforces Round #267 (Div. 2) B. Fedor and New Game【位运算/给你m+1个数让你判断所给数的二进制形式与第m+1个数不相同的位数是不是小于等于k,是的话就累计起来】

    After you had helped George and Alex to move in the dorm, they went to help their friend Fedor play ...

  9. Codeforces Round #267 (Div. 2) B. Fedor and New Game

    After you had helped George and Alex to move in the dorm, they went to help their friend Fedor play ...

随机推荐

  1. Drawable和Bitmap区别

    Bitmap - 称作位图,一般位图的文件格式后缀为bmp,当然编码器也有很多如RGB565.RGB888.作为一种逐像素的显示对象执行效率高,但是缺点也很明显存储效率低.我们理解为一种存储对象比较好 ...

  2. Spring配置AOP实现定义切入点和织入增强

    XML里的id=””记得全小写 经过AOP的配置后,可以切入日志功能.访问切入.事务管理.性能监测等功能. 首先实现这个织入增强需要的jar包,除了常用的 com.springsource.org.a ...

  3. [Phalcon] DI默认的服务

    代码,说明一切 namespace Phalcon\Di\FactoryDefault; use Phalcon\Di\Service; use Phalcon\Di\FactoryDefault; ...

  4. (Interface)接口特点

    接口是一种规范.只要一个类继承了一个接口,这个类就必须实现这个接口中所有的成员 为了多态. 接口不能被实例化.也就是说,接口不能new(不能创建对象) 接口中的成员不能加"访问修饰符&quo ...

  5. Dynamics AX 2012 R2 如何处理运行时间较长的报表

    当处理的数据量较多,逻辑比较复杂时,报表可能会超时.为了解决这个问题,Reinhard一直使用SrsReportDataProviderPreProcess来做预处理报表.它会在调用SSRS前,在AX ...

  6. Dynamics AX 2012 R2 通过数据源保存记录时触发的方法

    我们都知道,在窗体上保存记录时,会像在表上保存时一样,触发很多方法.这里Reinhard找到了一个流程图,看看都触发了哪些方法,并且这些方法是以怎样的顺序被触发的. 窗体上数据源的Validate() ...

  7. If you only do what you can do you'll never be more than you are now.

    If you only do what you can do you'll never be more than you are now. 如果你只是做你力所能及的事,你将不会有所进步:

  8. iOS开发中设置UITextField的占位文字的颜色,和光标的颜色

    在iOS开发中,对于很多初学者而言,很有可能碰到需要修改UITextField的占位文字的颜色,以及当UITextField成为第一响应者后光标的颜色,那么下面小编就介绍一下修改占位文字和光标的颜色. ...

  9. 实时发布到tomcat

    当我们在运用eclipse进行web项目的文件编写的时候,希望编写的东西可以实时的发布在tomcat的webapps文件夹之中,那么我们应该怎么做呢!下面就是操作方法:   1. 首先你的eclips ...

  10. FireMonkey 保存图片到JPG的方法 BMP转JPG

    习惯VCL的做法了,到了FireMonkey里面,好像查不到单独的JPEG单元了,不少朋友就郁闷如何处理JPG了,这么大件事,不可能没有处理方法的,具体就请看代码: uses FMX.Surfaces ...