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. OpenCV整体的模块架构

    之前啃了不少OpenCV的官方文档,发现如果了解了一些OpenCV整体的模块架构后,再重点学习自己感兴趣的部分的话,就会有一览众山小的感觉,于是,就决定写出这篇文章,作为启程OpenCV系列博文的第二 ...

  2. 基础-DP

    Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like ...

  3. Windows Phone 七、XML序列化

    DataContractSerializer对象 public class Person { public int Id { get; set; } public string Name { get; ...

  4. 最新IP地址数据库Dat格式-高性能高并发版(2017年1月)

    最新IP地址数据库->Dat格式 高性能格式->qqzeng-ip.dat 国内版-20170101-Dat 版                国外版-20170101-Dat 版     ...

  5. matlab画柱状图

    论文中需要画图进行比较,感觉还是matlab画起来比较方便,先把自己画的图及matlab代码放上. y=[300 311;390 425; 312 321; 250 185; 550 535; 420 ...

  6. Canvas: Out of system resources

    一个手写板的项目 在线程中操作Canvas画用户的笔记, 画不了几笔就卡住不画了, 然后保存到另外的image时 提示“Out of system Resource”错误, 百思不得姐 中间考虑是不是 ...

  7. AngularJs--angular-pagination可复用的分页指令

    1.angular-pagination 是基于angular 编写的可复用分页指令 安装 克隆项目到本地: git clone https://github.com/febobo/angular-p ...

  8. bitnami redmine每日自动备份

    主要思路:在半夜时停止服务,进行完整备份,然后再开启服务. 1.主脚本backup.bat: call backup-stopserver.batping /n 20 127.1 >nul ca ...

  9. s3c2440液晶屏驱动 (非内核自带) linux-4.1.24

    对于,不想逐一检查内核自带驱动,想自己编写驱动. 1,make menuconfig 去掉 编译到内核,改为 M 编译为 模块(因为要用到里面的3个.ko 驱动) Device Drivers --- ...

  10. Rest Post示例(java服务端、python客户端)

    前提:服务端是现成的,java.springMVC.resttemplate.jboss等:突然有个需要,要在windows上开发一个客户端,作用是定期向服务端上传文件.想了想,如果客户端写一个jav ...