题解分享
题解分享简介
士兵 - 题解
```
//数组模拟链表秒了
#include<bits/stdc++.h>
using namespace std;
int n,m;
struct soldier{
int index;
int pre;
int nxt;
};
int main(){
scanf("%d%d",&n,&m);
vector<soldier> s(n+10);
for(int i = 1; i <= n; i++){
s[i].index = i;
s[i].pre = i-1;
s[i].nxt = i+1;
}
int ind;
while(m--){
//不断维护每个死去士兵的左节点和右节点
scanf("%d",&ind);
int nx = s[ind].nxt;
int pr = s[ind].pre;
s[pr].nxt = s[ind].nxt;
s[nx].pre = s[ind].pre;
if(s[ind].nxt > n && s[ind].pre < 1) printf("* *\n");
else if(s[ind].nxt > n) printf("%d *\n",s[ind].pre);
else if(s[ind].pre < 1) printf("* %d\n",s[ind].nxt);
else printf("%d %d\n",s[ind].pre, s[ind].nxt);
}
}
```
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士兵 - 题解
```
#include <cstdio>
const int MAXN = 1000010;
int e[MAXN], l[MAXN], r[MAXN], idx;
int pos[MAXN]; // 记录每个士兵对应的节点索引
void init() {
r[0] = 1;
l[1] = 0;
idx = 2;
}
void insert(int a, int x) {
e[idx] = x;
l[idx] = a;
r[idx] = r[a];
l[r[a]] = idx;
r[a] = idx++;
}
void remove(int a) {
l[r[a]] = l[a];
r[l[a]] = r[a];
}
int main() {
int n, m;
scanf("%d%d", &n, &m);
init();
int last = 0; // 初始左端点
for (int i = 1; i <= n; ++i) {
insert(last, i);
pos[i] = idx - 1; // 当前插入的节点索引
last = pos[i]; // 更新为最新插入的节点
}
while (m--) {
int s;
scanf("%d", &s);
int a = pos[s];
int left = l[a], right = r[a];
if (left == 0) printf("* ");
else printf("%d ", e[left]);
if (right == 1) printf("*\n");
else printf("%d\n", e[right]);
remove(a); // 删除该士兵节点
}
return 0;
}
```
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