博主開始複習PAT甲級了,會做多個複習的專題,會解釋需要注意的基本概念和題庫相關的滿分代碼展示。
目的:通過多刷題,多總結回顧,讓自己在方法層面還是代碼細節達到熟練寫出的程度。
形式:思維導圖+代碼點評
提示:每個人的基礎不同,這裏不做普適性的建議。我出的專題重點是串聯知識,僅供參考。
Dijkstra專題近三年考察情況:19年3月第三題
1158之前寫過了:https://blog.csdn.net/allisonshing/article/details/104361040
放出來的都是我優化的liuchuo的源代碼,沒放出來的就按照她的寫法就行。
1107. Social Clusters (30)
https://pintia.cn/problem-sets/994805342720868352/problems/994805361586847744
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> father, isRoot;
int cmp1(int a, int b){return a > b;}
//我改了下面兩個函數,liuchuo版的是非遞歸的
int findFather(int x) {
return x == father[x]? x : (father[x]=findFather(father[x]));
}
void Union(int a, int b) {
father[findFather(a)]=findFather(b);
}
int main() {
int n, k, t, cnt = 0;
int course[1001] = {0};
scanf("%d", &n);
father.resize(n + 1);
isRoot.resize(n + 1);
for(int i = 1; i <= n; i++)
father[i] = i;
for(int i = 1; i <= n; i++) {
scanf("%d:", &k);
for(int j = 0; j < k; j++) {
scanf("%d", &t);
if(course[t] == 0)
course[t] = i;
Union(i, findFather(course[t]));
}
}
for(int i = 1; i <= n; i++)
isRoot[findFather(i)]++;
for(int i = 1; i <= n; i++) {
if(isRoot[i] != 0) cnt++;
}
printf("%d\n", cnt);
sort(isRoot.begin(), isRoot.end(), cmp1);
for(int i = 0; i < cnt; i++) {
printf("%d", isRoot[i]);
if(i != cnt - 1) printf(" ");
}
return 0;
}
1114 Family Property (25分)
https://pintia.cn/problem-sets/994805342720868352/problems/994805356599820288
1118 Birds in Forest (25分)
https://pintia.cn/problem-sets/994805342720868352/problems/994805354108403712