POJ_2502_建图+最短路

POJ 2502

  • 感觉题目意思不清晰,主要是建图,然后是最短路.
  • 首先将速度换算成分钟,将所有输入的点都离散一下,然后处理-1 -1与一条地铁路线之间的时间(双向).
  • 然后处理不同地铁路线之间点的距离 grid[i][j] = grid[j][i] = min(grid[i][j],dis(point[i],point[j])/v1) 
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#include <iostream>
#include <math.h>
#include <queue>
#include <string.h>

using namespace std;
const int maxn = 305;
const int MAX = 1<<30;

struct node {
	double x,y;
} point[maxn];

bool vis[maxn];
double grid[maxn][maxn];
double dist[maxn];
int n;

double dis(node a,node b) {
	double dx = a.x - b.x;
	double dy = a.y - b.y;
	return sqrt(dx*dx+dy*dy);
}
void spfa() {
	for (int i=1; i<=n; i++) dist[i] = MAX;
	memset(vis,0,sizeof vis);
	queue<int> que;
	vis[1] = 1; dist[1] = 0;
	que.push(1);
	while (!que.empty()) {
		int cur = que.front(); que.pop();
		vis[cur] = 0;
		for (int i=1; i<=n; i++) {
			if (dist[i] > dist[cur] + grid[cur][i]) {
				dist[i] = dist[cur] + grid[cur][i];
				if (!vis[i]) {
					vis[i] = 1;
					que.push(i);
				}
			}
		}
	}
}
int main() {
//	freopen("a.txt","r",stdin);
	double v1 = 10000.0/60; double v2 = 40000.0/60;
	cin >> point[1].x >> point[1].y >> point[2].x >> point[2].y;
	for (int i=1; i<maxn; i++) 
		for (int j=1; j<maxn; j++) {
			if (i == j) grid[i][j] = 0;
			else grid[i][j] = MAX;
		}
	double dx,dy;
	n = 2;
	int cnt = 3;
	while (scanf("%lf%lf",&dx,&dy) != EOF) {
		if (dx == -1 && dy == -1) {
			cnt = n+1;
			continue;
		}
		n++;
		point[n].x = dx; point[n].y = dy;
		if (cnt != n) { //给地铁建图
			grid[n][n-1] = grid[n-1][n] = min(grid[n][n-1],dis(point[n],point[n-1])/v2);
		}
	}
	for (int i=1; i<=n; i++) 
		for (int j=1; j<=n; j++) 
			grid[i][j] = grid[j][i] = min(grid[i][j],dis(point[i],point[j])/v1);
	spfa();
	printf("%.0lf\n",dist[2]);
	return 0;
}
算法

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