uva 718 - Skyscraper Floors(数论+bfs)

题目链接：uva 718 - Skyscraper
Floors

题目大意：一栋大楼，有F层楼，E个电梯，现在要从A层到B层，问是否可行，每个电梯给出Xi和Yi,代表这个电梯可以到达的层数Yi+k∗Xi(k≥0)

解题思路：建图，以A，B以及电梯为节点建图，将可以到达A，B这两层的电梯与这两点建边，在将两两电梯可以达到同一层的建边，判断方法为:Yi+aXi=Yj+bXj,移项得：aXi+bXj=Yj−Yi,即是一个线性方程，用拓展欧几里得算法求出通解的形式，判断是否存在通解在0~F之间即可。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>

using namespace std;
const int maxn = 105;
const int INF = 0x3f3f3f3f;
int F, E, A, B, X[maxn], Y[maxn];
vector<int> g[maxn];

void gcd (int a, int b, int& d, int& x, int& y) {
if (b == 0) {
d = a;
x = 1;
y = 0;
} else {
gcd(b, a%b, d, y, x);
y -= (a/b)*x;
}
}

void addPoint (int k, int u, int v) {
if (k < Y[v])
return ;
if ((k-Y[v]) % X[v] == 0) {
g[v].push_back(u);
g[u].push_back(v);
}
}

void addEdge (int u, int v) {
int a = X[u], b = -X[v], c = Y[v] - Y[u];
int d, xi, yi;
gcd(a, b, d, xi, yi);

if (c % d)
return ;

int lower = -INF, up = INF;

double T = (F - Y[u]) * 1.0 / X[u];
if (b / d > 0) {
up = min (up, (int)floor((T*d - xi*c) / b));
lower = max(lower, (int)ceil(-xi*c*1.0/b));
} else {
up = min(up, (int)floor(-xi*c*1.0/b));
lower = max(lower, (int)ceil((T*d - xi*c) / b));
}

T = (F - Y[v]) * 1.0 / X[v];
if (a / d > 0) {
up = min(up, (int)floor((yi*c*1.0)/a));
lower = max(lower, (int)ceil((T*d*-yi*c) / a));
} else {
up = min(up, (int)floor((T*d*-yi*c) / a));
lower = max(lower, (int)ceil((yi*c*1.0)/a));
}

if (up < lower)
return;
g[u].push_back(v);
g[v].push_back(u);
}

void init () {
scanf("%d%d%d%d", &F, &E, &A, &B);
for (int i = 0; i <= E+1; i++)
g[i].clear();

for (int i = 1; i <= E; i++) {
scanf("%d%d", &X[i], &Y[i]);
addPoint(A, 0, i);
addPoint(B, E+1, i);
}

for (int i = 1; i <= E; i++) {
for (int j = 1; j <= E; j++)
addEdge(i, j);
}
}

bool bfs (int s, int e) {
int vis[maxn];
memset(vis, 0, sizeof(vis));

queue<int> que;
que.push(s);
vis[s] = 1;

while (!que.empty()) {
int u = que.front();
que.pop();

if (u == e)
return true;

for (int i = 0; i < g[u].size(); i++) {
int v = g[u][i];

if (vis[v])
continue;

que.push(v);
vis[v] = 1;
}
}
return false;
}

int main () {
int cas;
scanf("%d", &cas);
while (cas--) {
init ();
if (bfs(0, E+1))
printf("It is possible to move the furniture.\n");
else
printf("The furniture cannot be moved.\n");
}
return 0;
}