POJ - 2502 Subway 【最短路】

版权声明：如需转载，记得标识出处 https://blog.csdn.net/godleaf/article/details/88943386

题目链接：http://poj.org/problem?id=2502

所有点用走路的速度连接，相邻站点用地铁速度连接，跑最短路。

注意输出用  %.0f，取其他精度会WA

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <cmath>

using namespace std;

const int Maxn = 210+10;
const int INF = 0x3f3f3f3f;
const double esp = 1e-6;

struct Edge {
	int v, next;
	double w;
} edge[Maxn*Maxn+2*Maxn];

int edge_cnt, h[Maxn];
double dis[Maxn];
bool vis[Maxn];

vector<pair<int, int> > node[Maxn];

void add(int u, int v, double w) {
	edge[edge_cnt].v = v;
	edge[edge_cnt].w = w;
	edge[edge_cnt].next = h[u];
	h[u] = edge_cnt++;
}

void spfa(int n) {
	memset(vis, false, sizeof(vis));
	for(int i = 0; i <= n; ++i) dis[i] = INF;
	dis[1] = 0; vis[1] = true;
	stack <int> qu;
	qu.push(1);

	while(!qu.empty()) {
		int u = qu.top(); qu.pop();
		vis[u] = false;

		for(int i = h[u]; i != -1; i = edge[i].next) {
			Edge &e = edge[i];
			if(fabs(dis[e.v]-(dis[u]+e.w)) > esp && dis[e.v] > dis[u]+e.w) {
				dis[e.v] = dis[u]+e.w;
				if(!vis[e.v]) {
					vis[e.v] = true;
					qu.push(e.v);
				}
			}
		}
	}
}

double dist(int x, int y, int xx, int yy) {
	return sqrt((double)(x-xx)*(x-xx)+(y-yy)*(y-yy));
}

bool input(int &cnt, int &n) {
	int x, y;
	while(1) {
		if(scanf("%d%d", &x, &y) == EOF) return false;
		if(x == -1 && y == -1) return true;
		node[cnt].push_back(make_pair(x, y));
		n++;
	}
}

int main(void)
{
	int m = 0, x1, x2, y1, y2, n = 2;
	scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
	while(input(m, n)) m++;
	map<pair<int, int>, int> num;
	num[make_pair(x1, y1)] = 1;
	num[make_pair(x2, y2)] = n;
	memset(h, -1, sizeof(h));
	edge_cnt = 0;

	int u, v, N = 2, x, y, xx, yy;
	for(int i = 0; i < m; ++i) {
		for(int j = 0; j < node[i].size(); ++j) {
			for(int k = 0; k < m; ++k) {
				for(int l = 0; l < node[k].size(); ++l) {
					if(i == k && l == j) continue;
					x = node[i][j].first; y = node[i][j].second;
					xx = node[k][l].first; yy = node[k][l].second;
					if(!num[make_pair(x, y)]) num[make_pair(x, y)] = N++;
					if(!num[make_pair(xx, yy)]) num[make_pair(xx, yy)] = N++;
					u = num[make_pair(x, y)];
					v = num[make_pair(xx, yy)];
					double d = dist(x, y, xx, yy);
					if(i == k && j+1 == l) {
						add(u, v, d*60/40000.0);
						add(v, u, d*60/40000.0);
					} else add(u, v, d*60/10000.0);
				}
			}
		}
	}
	for(int i = 0; i < m; ++i) {
		for(int j = 0; j < node[i].size(); ++j) {
			x = node[i][j].first; y = node[i][j].second;
			double d = dist(x1, y1, x, y);
			v = num[make_pair(x, y)];
			add(1, v, d*60/10000.0);
			add(v, 1, d*60/10000.0);
			d = dist(x2, y2, x, y);
			add(n, v, d*60/10000.0);
			add(v, n, d*60/10000.0);
		}
	}
	double d = dist(x1, y1, x2, y2);
	add(1, n, d*60/10000.0);
	add(n, 1, d*60/10000.0);
	spfa(n);
	printf("%.0f\n", dis[n]);
	return 0;
 }