Daliy Algorithm (GPLT)-- day 94
Nothing to fear
种一棵树最好的时间是十年前，其次是现在！
那些你早出晚归付出的刻苦努力，你不想训练，当你觉的太累了但还是要咬牙坚持的时候，那就是在追逐梦想，不要在意终点有什么，要享受路途的过程，或许你不能成就梦想，但一定会有更伟大的事情随之而来。 mamba out~
2020.7.6 - 7.7
人一我十，人十我百，追逐青春的梦想，怀着自信的心，永不言弃！
GPLT - L2-020 功夫传人
坑
- 只统计得道者得功夫值！读题要仔细
- 祖师爷也可能是得道者
- printf("%.f",ans) 会自动进行四舍五入，而题目要求只输出整数部分，故只能利用强制类型转换。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <vector>
using namespace std;
const int N = 100005;
int n;
double z , r;
vector<int> a[N];
double tot;
int dedao[N];
void dfs(int now,double val)
{
if(dedao[now]){
tot += val * dedao[now];
}
for(int i = 0;i < a[now].size(); i++)
{
dfs(a[now][i] ,val - val * r * 0.01);
}
}
int main()
{
cin >> n >> z >> r;
int k = 0;
for(int i = 0;i < n ;i ++)
{
cin >> k;
if(k == 0){
scanf("%d",&dedao[i]);
}else{
int x = 0;
for(int j = 0;j < k;j ++)
{
scanf("%d", &x);
a[i].push_back(x);
}
}
}
dfs(0 , z);
printf("%d\n",(int)tot);
return 0;
}
GPLT - L2-021 点赞狂魔
分析：
利用set记录不同得标签出现得次数，标签得平均出现次数为总数量 / 不同标签得数目。然后按照题目规则进行排序即可。
#include <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <algorithm>
#include <set>
using namespace std;
const int N = 505;
struct node{
string name;
int cnt;
int k;
}a[N];
bool cmp(node a ,node b)
{
if(a.cnt == b.cnt)
{
return (a.k * 1.0 / a.cnt) < (b.k * 1.0 / b.cnt);
}else return a.cnt > b.cnt;
}
int main()
{
int n;
cin >> n;
set<int> s;
for(int i = 0;i < n ;i ++)
{
cin >> a[i].name;
cin >> a[i].k;
for(int j = 0;j < a[i].k; j++)
{
int x = 0;
scanf("%d" ,&x);
s.insert(x);
}
a[i].cnt = s.size();
s.clear();
}
sort(a , a + n , cmp);
for(int i = 0;i < 3;i ++)
{
if(a[i].k!=0)
{
if(i == 0)
cout << a[i].name;
else cout << " " << a[i].name;
}else{
if(i == 0)
cout << "-";
else cout << " -";
}
}
return 0;
}
GPLT - L2-022 重排链表
没认真读题还以为要排序呢。。。，注意链表去重，老玩家了
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
using namespace std;
const int N = 1000005;
struct node{
int add , val , next;
}a[N] , Link[N] ,ans[N];
int head , n , len = 0;
bool cmp(node a,node b)
{
return a.val < b.val;
}
int main()
{
cin >> head >> n;
int x , y , z;
for(int i = 0;i < n;i ++)
{
scanf("%d %d %d",&x , & y, &z);
a[x] = {x , y , z};
}
for(int i = head ;i != -1 ;i = a[i].next)
{
Link[len++] = a[i];
}
int i = 0 , j = len - 1 ,id = 0;
while(i < j)
{
ans[id++] = Link[j--];
ans[id++] = Link[i++];
if(i == j){
ans[id++] = Link[i];
break;
}
}
for(int i = 0;i < len ;i ++)
{
if(i == len - 1)
{
printf("%05d %d -1\n",ans[i].add , ans[i].val);
}else{
printf("%05d %d %05d\n",ans[i].add , ans[i].val ,ans[i+1].add);
}
}
return 0;
}
GPLT - L2-023 图着色问题
PAT叫我做人
题目只说了图是合法得，没说图是一定联通得，所以要遍历每一个未被访问得结点，不然数据点3是过不去得 惨遭毒手
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <set>
#include <cstring>
using namespace std;
const int N = 250005;
const int M = 505;
struct edge{
int next;
int to;
}e[N];
int head[M],tot;
int n, m, k;
int color[M];
bool vis[M];
void add(int a,int b)
{
e[tot].to = b;
e[tot].next = head[a];
head[a] = tot++;
}
set<int> s;
void checkColor(int now, bool &flag)
{
vis[now] = true;
s.insert(color[now]);
for(int i = head[now] ; i ; i = e[i].next)
{
int next = e[i].to;
if(color[now] == color[next]){
flag = true;
return;
}
if(!vis[next] && !flag)
{
checkColor(next , flag);
}
}
}
int main()
{
cin >> n >> m >> k;
int x , y;
for(int i = 0;i < m;i ++)
{
scanf("%d %d",&x , &y);
add(x , y);add(y , x);
}
int q;
cin >> q;
for(int i = 0;i < q;i ++)
{
for(int j = 1;j <= n;j ++){
scanf("%d",&color[j]);
}
memset(vis , 0 , sizeof vis);
bool flag = false;
s.clear();
for(int i = 1;i <= n ;i ++)
{
if(!vis[i] && !flag)
{
checkColor(i, flag);
}
}
if(!flag && s.size() == k)printf("Yes\n");
else printf("No\n");
}
return 0;
}
GPLT - L2-024 部落
并查集模板题
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <vector>
using namespace std;
const int N = 10005;
int f[N] , cnt[N];
bool vis[N];
int find(int k)
{
if(k == f[k])return k;
else return f[k] = find(f[k]);
}
int Unoin(int a,int b)
{
int x = find(a) , y = find(b);
f[x] = y;
}
int main()
{
for(int i = 1;i <= N;i ++)f[i] = i;
int n , q;
cin >> n;
for(int i = 0;i < n;i ++)
{
int k = 0;cin >> k;
for(int j = 0;j < k;j ++)
{
cin >> cnt[j];
vis[cnt[j]] = 1;
}
for(int j = 1;j < k;j ++)
Unoin(cnt[j-1],cnt[j]);
}
int people = 0 , ans = 0;
for(int i = 1;i <= N;i ++)
{
if(vis[i] == 1){
if(i == f[i])ans++;
people++;
}
}
cout << people << " " << ans << endl;
cin >> q;
int a , b;
for(int i = 0;i < q;i ++)
{
cin >> a >> b;
int x = find(a) , y = find(b);
if(x != y)printf("N\n");
else printf("Y\n");
}
return 0;
}
GPLT - L2-025 分而治之
简单dfs判断连通分量即可，话说 说好的m不超过10000呢
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <cstring>
using namespace std;
const int N = 100005;
struct edge{
int to;
int next;
}e[N];
int head[N] , tot;
int n , m;
bool vis[N];
void add(int a , int b)
{
e[tot].to = b;
e[tot].next = head[a];
head[a] = tot++;
}
bool dfs(int x)
{
for(int i = head[x] ; i ; i = e[i].next)
{
int next = e[i].to;
if(vis[next] == false)return false;
}
return true;
}
int main()
{
cin >> n >> m;
int x , y;
for(int i = 0;i < m;i ++)
{
cin >> x >> y;
if(x == y)continue;
add(x , y) ,add(y , x);
}
int k , np;
cin >> k;
for(int time = 0;time < k;time ++)
{
cin >> np;
memset(vis , 0 , sizeof vis);
for(int i = 0;i < np;i ++)
{
cin >> x;
vis[x] = 1;
}
bool flag = true;
for(int i = 1;i <= n;i ++)
{
if(!vis[i])
{
if(!dfs(i))
{
flag = false;
break;
}
}
}
if(flag)printf("YES\n");
else printf("NO\n");
}
return 0;
}
