NYOJ 511 - 移动小球 链表版

传送门NYOJ 511 - 移动小球 链表版

╮(╯▽╰)╭本来昨天要干的,可是昨天莫名的烦躁,于是就听了一个晚上的歌╮(╯▽╰)╭

刚开始写时没想到用结构体数组,..于是就各种情况各种头指针...

后来参考了一下别人的才发现可以用结构体数组来做...

这也是第一次学习双向链表,这题做完后感觉我已经差不多可以掌握了...╮(╯▽╰)╭

详情见代码 I

--------------------------------------------14. 5. 6补充.----------------------------------------------------

一个半月后重新看这题, 发现可以用STL实现, 代码将会更加简洁... 晚上回来再搞.

用STL的优点: 代码简单, 不容易出错

缺点: 由于list中查找元素的时间复杂度是O(n), 比不上用结构体数组的O(1), 所以所用时间会大大增加. 大家自己取舍.

详情见代码 II

-------------------------------------------------------------------------------------------------------------

#include <stdio.h>struct ball{int num;struct ball *left, *right;}ball[11000];void Initial(int n);void Leftmove(int x, int y);void Rightmove(int x, int y);int main(){//freopen("input.txt","r",stdin);int T;scanf("%d",&T);while (T--){int n, m;char c;scanf("%d%d", &n, &m);Initial(n);while (m--){char c;int x, y;getchar();scanf("%c%d%d",&c, &x, &y);if (c == 'A')Leftmove(x, y);else if (c == 'B')Rightmove(x, y);elseif (x == 1)printf("%d\n",ball[y].right->num);elseprintf("%d\n",ball[y].left->num);}}return 0;}void Initial(int n){for (int i = 1; i <= n; i++){ball[i].num = i;ball[i].left = &ball[i - 1];ball[i].right = &ball[i + 1];}ball[1].left = &ball[n];ball[n].right = &ball[1];}void Leftmove(int x, int y){ball[x].left->right = ball[x].right;ball[x].right->left = ball[x].left;ball[x].right = &ball[y];ball[x].left = ball[y].left;ball[x].left->right = &ball[x];ball[y].left = &ball[x];}void Rightmove(int x, int y){ball[x].left->right = ball[x].right;ball[x].right->left = ball[x].left;ball[y].right->left = &ball[x];ball[x].right = ball[y].right;ball[x].left = &ball[y];ball[y].right = &ball[x];}

代码 II

#include <bits/stdc++.h>using namespace std;int main(){//freopen("input.txt", "r", stdin);    list<int> ball;list<int>::iterator itNow, itTarget;    int T, n, i, num, now, target;char ch;    scanf("%d", &T);    while (T--)    {ball.clear();        scanf("%d%d%*c", &num, &n);        for (i = 1; i <= num; i++)ball.push_back(i);        while (n--)        {ch = getchar();scanf("%d%d%*c", &now, &target);itNow = find(ball.begin(), ball.end(), now);itTarget = find(ball.begin(), ball.end(), target);if (ch == 'A'){ball.insert(itTarget, now);ball.erase(itNow);}else if (ch == 'B'){ball.insert(++itTarget, now);ball.erase(itNow);}else//1询问右边的, 0左边.{if (now == 1 && itTarget == --ball.end())printf("%d\n", *ball.begin());else if (now == 0 && itTarget == ball.begin())printf("%d\n", *(--ball.end()));else if (now == 1)printf("%d\n", *(++itTarget));else if (now == 0)printf("%d\n", *(--itTarget));}}}return 0;}