UVA 718 - Skyscraper Floors(數論)

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UVA 718 - Skyscraper Floors

題目連結

題意：在一個f層高的樓上，有e個電梯，每個電梯有x，y表示y + k * x層都可以到，現在要問從a層能否到達b層（中間怎麼轉乘電梯不限制）

思路：對於兩個電梯間能不能轉乘，只要滿足y[i] + xx x[i] == y[j] + yy y[j].然後移項一下，就可以用拓展歐幾裡得求解，進而求出x,y的通解，然後利用通解範圍x‘ >= 0, y‘ >= 0, x[i]x‘ + y[i] <= f, x[j] y‘ + y[j] <= f,求出通解中t的上限和下限，就可以判斷有無解了，然後就建圖，把兩個可以轉乘電梯建邊，然後在讓a, b分別和電梯判斷能不能到，能到建邊，最後dfs一遍就能判斷能否達到了。

代碼：

#include <stdio.h>#include <string.h>#include <vector>#include <math.h>using namespace std;const int INF = 0x3f3f3f3f;int t, f, e, a, b, x[105], y[105], vis[105];vector<int> g[105];int exgcd(int a, int b, int &x, int &y) {if (!b) {x = 1; y = 0; return a;}int d = exgcd(b, a % b, y, x);y -= a / b * x;return d;}void build(int v, int u, int i) {if (v < y[i]) return;if ((v - y[i]) % x[i] == 0) {g[u].push_back(i);g[i].push_back(u); }}void build2(int i, int j) {int xx, yy;int a = x[i], b = -x[j], c = y[j] - y[i];int d = exgcd(a, b, xx ,yy);if (c % d) return;int down = -INF;int up = INF;if (b / d > 0) { down = max(down, (int)ceil(-xx * c * 1.0 / b)); up = min(up, (int)floor(((f - y[i]) * 1.0 * d / x[i] - xx * c * 1.0) / b));}else {down = max(down, (int)ceil(((f - y[i]) * 1.0 * d / x[i] - xx * c * 1.0) / b)); up = min(up, (int)floor(-xx * c * 1.0 / b));}if (a / d > 0) {down = max(down, (int)ceil((yy * c * 1.0 - (f - y[j]) * 1.0 * d / x[j]) / a)); up = min(up, (int)floor(yy * c * 1.0 / a));}else { down = max(down, (int)ceil(yy * c * 1.0 / a)); up = min(up, (int)floor((yy * c * 1.0 - (f - y[j]) * 1.0 * d / x[j]) / a));}if (down > up) return;g[i].push_back(j);g[j].push_back(i);}bool dfs(int u) {if (u == e + 1) return true;vis[u] = 1;for (int i = 0; i < g[u].size(); i++) {int v = g[u][i];if (vis[v]) continue;if (dfs(v)) return true; } return false;}int main() {scanf("%d", &t);while (t--) {memset(g, 0, sizeof(g));scanf("%d%d%d%d", &f, &e, &a, &b);for (int i = 1; i <= e; i++) {scanf("%d%d", &x[i], &y[i]);   build(a, 0, i);   build(b, e + 1, i);}for (int i = 1; i <= e; i++) {for (int j = i + 1; j <= e; j++) {build2(i, j);   }  }  memset(vis, 0, sizeof(vis));  if (dfs(0)) printf("It is possible to move the furniture.\n");  else printf("The furniture cannot be moved.\n"); }return 0;}