首页 > 代码库 > UVA 1444 - Knowledge for the masses
UVA 1444 - Knowledge for the masses
UVA 1444 - Knowledge for the masses
题目链接
题意:给定R排书架,现在要求最小代价移动书架打开一条通道,每次移动书架的代价为1(距离不限),问最小代价和最小代价的位置
思路:对于每一行,从左往右再从右往左各推一遍,每次把各个位置代价的最小值算出来,计算的过程要记录每个位置对应前面空位个数和空位位置,这样每个书架要移动的代价就能快速算出,最后处理往后,在每个位置遍历找出最小值位置即可
代码:
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int INF = 0x3f3f3f3f; const int N = 1000005; int t, R, L, n, blo[N], zero[N], value[N], can[N], g[N]; void build() { scanf("%d", &n); int zn = 0, now = 0; memset(value, -1, sizeof(value)); for (int i = 0; i < n; i++) { scanf("%d", &blo[i]); if (blo[i] == 0) { zero[zn++] = i; can[now]++; now++; } else { now += blo[i]; int k = min(zn, blo[i]); for (int j = 1; j <= k; j++) { value[now - j] = i - zero[zn - j] - j + 1; can[now - j]++; g[now - j] += value[now - j]; } } } reverse(blo, blo + n); zn = 0; now--; for (int i = 0; i < n; i++) { if (blo[i] == 0) { zero[zn++] = i; now--; } else { now -= blo[i]; int k = min(zn, blo[i]); for (int j = 1; j <= k; j++) { if (value[now + j] == -1) { value[now + j] = i - zero[zn - j] - j + 1; g[now + j] += value[now + j]; can[now + j]++; } else { int tmp = i - zero[zn - j] - j + 1; g[now + j] += min(0, tmp - value[now + j]); } } } } } void solve() { int an = 0, Min = INF; for (int i = 0; i < L; i++) { if (can[i] != R) continue; if (Min > g[i]) { Min = g[i]; an = 0; value[an++] = i; } else if (Min == g[i]) { value[an++] = i; } } printf("%d\n", Min); for (int i = 0; i < an; i++) printf("%d ", value[i]); printf("\n"); } int main() { scanf("%d", &t); while (t--) { scanf("%d%d", &R, &L); memset(can, 0, sizeof(can)); memset(g, 0, sizeof(g)); for (int i = 0; i < R; i++) build(); solve(); } return 0; }
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。