拼了老命用一种贪心的思想把它A了,但是代码写的太烂了,而且时间复杂度为 n ^ 2,我就不多说了,太烂了
之后上网找了一个规律,时间复杂度为 nlogn,而且思路很明确,又写了一遍
代码:(贪心)
/*
ID: 18906421
LANG: C++
PROG: sort3
*/
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn = 1005;
int n;
int arr[maxn];
int sorted[maxn];
int vis[4] = {0};
int s[3];
int find_pos(int u,int v,int num){
int re_pos;
for(int i = 0; i < num; i++){
if(arr[i] == v){
if(i >= s[u])
return i;
else
re_pos = i;
}
}
return re_pos;
}
int main(){
int n,cnt = 0,pos;
freopen("sort3.in","r",stdin);
freopen("sort3.out","w",stdout);
scanf("%d",&n);
for(int i = 0; i < n; i ++){
scanf("%d",&arr[i]);
sorted[i] = arr[i];
vis[arr[i]]++;
}
s[1] = -1; s[2] = -1;
sort(sorted,sorted + n);
for(int i = 0; i < n; i++)
if(sorted[i] == 1 && s[1] < 0) s[1] = i;
else if(sorted[i] == 2 && s[2] < 0) s[2] = i;
//printf("%d %d\n",s[1],s[2]);
for(int i = n - 1; i >= 0; i--){
//printf("%d %d %d\n",vis[1],vis[2],vis[3]);
if(arr[i] == 1){
if(vis[3]){
pos = find_pos(1,3,i);
//printf("%d %d\n",i,pos);
arr[pos] = 1;
cnt ++;
vis[3] --;
}
else if(vis[2]){
pos = find_pos(1,2,i);
//printf("%d %d\n",i,pos);
arr[pos] = 1;
cnt ++;
vis[2] --;
}
else
vis[1] --;
}
else if(arr[i] == 2){
if(vis[3]){
pos = find_pos(2,3,i);
//printf("%d %d\n",i,pos);
arr[pos] = 2;
cnt ++;
vis[3] --;
}
else vis[2] --;
}
else
vis[3] --;
}
printf("%d\n",cnt);
return 0;
}
思路:
/*
ID: 18906421
LANG: C++
PROG: sort3
*/
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn = 1005;
int n,ans;
int arr[maxn];
int sorted[maxn];
int vis[4][4] = {0};
int main(){
freopen("sort3.in","r",stdin);
freopen("sort3.out","w",stdout);
int n;
scanf("%d",&n);
for(int i = 0; i < n; i++){
scanf("%d",&arr[i]);
sorted[i] = arr[i];
}
sort(sorted,sorted + n);
for(int i = 0; i < n; i++){
if(arr[i] != sorted[i])
vis[sorted[i]][arr[i]] ++;
}
int v1 = min(vis[1][2],vis[2][1]);
vis[1][2] -= v1; vis[2][1] -= v1;
int v2 = min(vis[1][3],vis[3][1]);
vis[1][3] -= v2; vis[3][1] -= v2;
int v3 = min(vis[2][3],vis[3][2]);
vis[2][3] -= v3; vis[3][2] -= v3;
ans = v1 + v2 + v3;
ans += (vis[1][2] + vis[2][1] + vis[1][3] + vis[3][1] + vis[2][3] + vis[3][2]) / 3 * 2;
printf("%d\n",ans);
return 0;
}