PAT甲级1013-1014-1015

题目：1013 Battle Over Cities

思路：城市数也就1000， 对于每次询问暴力bfs一下看一下有多少连通块就行了。答案就是联通块数减一。

#include<stdio.h>
#include<stdlib.h>
#include<map>
#include<set>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<cmath> 
#include<queue>

#define inf 0x7fffffff
using namespace std;
typedef long long LL;
typedef pair<int, int> pr;

int n, m, k;
const int maxm = 1e6 + 5;
const int maxn = 1005;
struct edge{
    int u, v, nxt;
}e[maxm * 2];
int head[maxn], tot = 1;

void addedge(int u, int v)
{
    e[tot].u = u;
    e[tot].v = v;
    e[tot].nxt = head[u];
    head[u] = tot++;
    e[tot].u = v;
    e[tot].v = u;
    e[tot].nxt = head[v];
    head[v] = tot++;    
} 
 
int vis[maxn];
void bfs(int c, int city)
{
    queue<int>que;
    que.push(c);
    vis[c] = true;
    while(!que.empty()){
        int now = que.front();que.pop();
        for(int ed = head[now]; ed != -1; ed = e[ed].nxt){
            if(e[ed].v != city && !vis[e[ed].v]){
                vis[e[ed].v] = true;
                que.push(e[ed].v);
            }
        }
    }
    return ;
}

int main()
{
    scanf("%d%d%d", &n, &m, &k);
    memset(head, -1, sizeof(head));
    for(int i = 0; i < m; i++){
        int u, v;
        scanf("%d%d", &u, &v);
        addedge(u, v);
    }
    for(int i = 0; i < k; i++){
        int city;
        scanf("%d", &city);
        memset(vis, 0, sizeof(vis));
        int cnt = 0;
        for(int i = 1; i <= n; i++){
            if(!vis[i] && i != city){
                bfs(i, city);
                cnt++;
            }
        }
        printf("%d
", cnt - 1);
    }
    return 0;
}

题目：1014 Waiting in Line

思路：大模拟。我好菜系列。

用队列模拟每个窗口排队的人。没满的时候就是从左到右排就行了，满了之后就是哪个队伍先有人走哪个队伍就先进去。这里用优先队列模拟。

坑点是，如果在五点之前被服务了，服务时间超过五点也没关系，但是如果开始服务时间就已经超过五点了就要Sorry。

还有一个地方写错了是刚开始所有人都排进去就直接结束了，但是队列里其实还有人。

#include<stdio.h>
#include<stdlib.h>
#include<map>
#include<set>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<cmath> 
#include<queue>

#define inf 0x7fffffff
using namespace std;
typedef long long LL;
typedef pair<int, int> pr;

int n, m, k, q;
const int maxn = 1005;
int process[maxn], total[maxn];
queue<int>que[25];
int t[25];

struct node{
    int time, line;
    node(){
    }
    node(int t, int l)
    {
        time = t;
        line = l;
    }
    bool operator < (const node& a)const{
        if(time == a.time)return line > a.line; 
        return time > a.time;
    }
};

int main()
{
    scanf("%d%d%d%d", &n, &m, &k, &q);
    for(int i = 1; i <= k; i++){
        scanf("%d", &process[i]);
    }
    
    int id = 1;
    priority_queue<node>lineque;
    for(int i = 1; i <= m; i++){
        for(int j = 1; j <= n; j++){
            que[j].push(id++);
            if(id > k)break;
        }
        if(id > k)break;
    }
    
    //que[1].push(id++);
    
//    for(int i = 1; i <= n; i++){
//        lineque.push(node(0, i));
//    }
    int sum = k;
    while(id <= k){
        //int linetime = 10000, line;
        for(int i = 1; i <= n; i++){
            if(!que[i].empty()){
                int peo = que[i].front();que[i].pop();
                sum--;
                t[i] += process[peo];
                total[peo] = t[i];
                lineque.push(node(total[peo], i));
            }
//            if(linetime > t[i]){
//                linetime = t[i];
//                line = i;
//            }
        }
//        que[line].push(id++);
        
        node l = lineque.top();lineque.pop();
        que[l.line].push(id++);
    }
    
    while(sum){
        for(int i = 1; i <= n; i++){
            if(!que[i].empty()){
                int peo = que[i].front();que[i].pop();
                sum--;
                t[i] += process[peo];
                total[peo] = t[i];
            }
        }
    }
    
    for(int i = 0; i < q; i++){
        int p;
        scanf("%d", &p);
        int h = total[p] / 60;
        int min = total[p] % 60;
        //printf("%d:%d
",p,total[p]);
        if(h + 8 >= 17 && min != 0 && total[p] - process[p] >= 540){
            printf("Sorry
");
        }
        else{
            printf("%02d:%02d
", h + 8, min);
        }
    }
    return 0;
}

题目：1015 Reversible Primes

思路：感觉PAT的题意每次讲的都好不清楚啊【远没有ACM表述清晰严谨】。

题目的意思是一个数10进制下是质数，然后转换成D进制再逆序之后的字符串当成10进制也是质数。

#include<stdio.h>
#include<stdlib.h>
#include<map>
#include<set>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<cmath> 
#include<stack>
#include<queue>

#define inf 0x7fffffff
using namespace std;
typedef long long LL;
typedef pair<int, int> pr;

int n, d;
const int maxn = 1e6 + 6;
bool p[maxn];

void isprime()
{
    p[0] = true;
    p[1] = true;
    for(int i = 2; i <= maxn; i++){
        if(p[i])continue;
        int j = 2;
        while(j * i <= maxn){
            p[j * i] = true;
            j++;
        }
    }
}

//stack<int>sss;
int getnum(int k)
{
    int res = 0, r = 1;
    int tmp = n;
    while(tmp){
        res += r * (tmp % k);
        //sss.push(tmp % k);
        tmp /= k;
        r *= k;
    }
    return res;
}

int getreverse(int k)
{
    int res = 0, tmp = n;
    while(tmp){
        res = res * k + tmp % k;
        tmp /= k;
    }
    return res;
}

int main()
{
    isprime();
       while(scanf("%d", &n) != EOF && n >= 0){
           scanf("%d", &d);
           //cout<<getnum()<<endl<<getreverse()<<endl;
        if(!p[getnum(10)] && !p[getreverse(d)]){
            printf("Yes
");
        }     
        else{
            printf("No
");
        }
    }
       
    return 0;
}