uva 12657 移动盒子

来源：互联网发布：提取伴奏的软件编辑：程序博客网时间：2024/05/17 22:07

这道题目采用双向链表数据结构，用left[i]和right[i]分别表示编号为i的盒子左边和右边的盒子编号，对于不同指令变换为对left和right数组的操作。最后遍历一个数组即可。题目中细节操作较多，需要细心检查。。。

#include<cstdio>#include<cstdlib>#include<algorithm>using namespace std;#define MAXL 100010int right[MAXL], left[MAXL];inline void link(int l, int r) {right[l] = r;left[r] = l;}int main() {//freopen("input.txt","r",stdin);int n, m, cases = 1;while(scanf("%d%d", &n, &m) != EOF) {int i, sta = 0;right[n] = 0;left[0] = n;for(i = 0; i < n; i++) {link(i, i+1);}for(i = 1; i <= m; i++) {int p, x, y;scanf("%d", &p);if(p == 4) sta = !sta;else{scanf("%d%d", &x, &y);if(p == 3 && right[y] == x) swap(x, y);if(sta && (p == 1 || p == 2)) p = 3 - p;if(p == 1 && right[x] == y) continue;if(p == 2 && right[y] == x) continue;}int lx = left[x], rx = right[x], ly = left[y], ry = right[y];switch (p) {case 1 : {link(lx, rx);link(ly, x);link(x, y);break;}case 2 : {link(lx, rx);link(x, ry);link(y, x);break;}case 3 : {link(lx, y);link(x, ry);if(rx == y) link(y, x);else{link(y, rx);link(ly, x);} break;}default : break;}}long long ans = 0;int b = 0;for(i = 1; i <= n; i++) {b = right[b];if(i % 2 == 1) ans += b;}if(sta && n % 2 == 0) ans = (long long)n * (n + 1) / 2 - ans;printf("Case %d: %lld\n", cases++, ans);}return 0;}

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