POJ 2502 Subway 【最短路之建图为关键】

传送门

题意: 有n条地铁线, 相邻两站可以做地铁到达, 速度为40km/h, 可以在任意站换乘其他地铁线必须走路， 速度为10km/h, 给定起点和终点问最少需要多少分钟才能到终点.

思路: 很明显这是一个最短路, 关键在于建图, 因为每个点都要向终点连一条速度为10km/h的边, 同时起点也想每个站连边, 同一条地铁线的相邻站连40km/h的边, 不同的地铁线中的站也要两两连走路的边, 然后跑最短路即可. 具体建图过程请看代码实现.

AC Code

const int maxn = 3e2+5;
const int maxm = 2e5 + 5;
struct node
{
int to, next; db w;
bool operator < (const node& a) const {
return w > a.w;
}
} e[maxm];

int cnt, head[maxn];
void add(int u, int v, db w) {
e[cnt] = node{v, head[u], w};
head[u] = cnt++;
}

void init() {
cnt = 0;
Fill(head, -1);
}

bool vis[maxn];
db dis[maxn];
int id;
void dij(int st,int ed)
{
priority_queue<node> q;
for (int i = 1 ; i <= id ; i ++) dis[i] = inf;
Fill(vis,0); dis[st] = 0;
q.push(node{st, 0, 0});
while (!q.empty()) {
node u = q.top();
q.pop();
if (u.to == ed) break;
if(vis[u.to]) continue;
vis[u.to] = 1;

for (int i = head[u.to]; ~i; i = e[i].next) {
int to = e[i].to;
if (dis[to] > dis[u.to] + e[i].w) {
dis[to] = dis[u.to] + e[i].w;
q.push(node{to, 0, dis[to]});
}
}
}
printf("%.0f\n", dis[ed]);
}
const db walkv = 500.0 / 3;
const db subv = 2000.0 / 3;
struct point {
db x, y;
}pp[maxn];
db cal(point a, point b) {
return sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y));
}
void solve()
{
init();
scanf("%lf%lf%lf%lf", &pp[1].x, &pp[1].y, &pp[2].x, &pp[2].y);
int flag = 1;  id = 2; db tx, ty;
while(cin >> tx >> ty) {
if (tx == -1 && ty == -1) {
flag = 1; continue;
}
pp[++id].x = tx; pp[id].y = ty;
if (flag > 1) {  // 相邻地铁线
add(id-1, id, cal(pp[id-1], pp[id])/subv);
add(id, id-1, cal(pp[id-1], pp[id])/subv);
}
flag++;
}
for (int i = 1 ; i <= id ; i ++) { // 所以点之间两两连线.
for (int j = i + 1 ; j <= id ; j++) {
add(i, j, cal(pp[i], pp[j])/walkv);
add(j, i, cal(pp[j], pp[i])/walkv);
}
}
dij(1, 2);
}