HDOJ 1245 Saving James Bond

题目链接：............

思路：岛为起点，把每个两个坐标距离小于人能跳的距离加入图中（用邻接矩阵存），点到岸边的距离小于人能跳得距离也加入图中，就成了最短路问题（这里用dijkstra算法)。

code:

#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <string.h>
typedef struct
{
	int v, next;
	double w;
}node;
node edge[40000];
int n = 0, k = 0, coor[105][2], head[102], used[102], step[102];
double dist[10200], d = 0;
void add(int u, int v, double w)
{
	edge[k].v = v;
	edge[k].w = w;
	edge[k].next = head[u];
	head[u] = k++;
}
void dijkstar()
{
	int i = 0, j = 0, cur = 0, l = 0;
	double min = 0;
	for(i = 0; i<=n+1; i++)
		dist[i] = 1234567890;
	for(cur = head[0]; cur != -1; cur = edge[cur].next)
	{
		step[edge[cur].v] = 1;
		dist[edge[cur].v] = edge[cur].w;
	}
	for(i = 0; i<=n+1; i++)
	{
		min = 1234567890;
		for(j = 0; j<=n+1; j++)
			if(min>dist[j] && !used[j])
			{
				min = dist[j];
				l = j;
			}
		used[l] = 1;
		for(cur = head[l]; cur != -1; cur = edge[cur].next)
		{
			if(!used[edge[cur].v] && dist[edge[cur].v]>dist[l]+edge[cur].w)
			{
				dist[edge[cur].v] = dist[l]+edge[cur].w;
				step[edge[cur].v] = step[l]+1;
			}
			else if(dist[edge[cur].v] == dist[l]+edge[cur].w && step[edge[cur].v] > step[l]+1)//到达这点的距离相同，选步数少的那个
				step[edge[cur].v] = step[l]+1;
		}
	}
}
int main()
{
	int i = 0, j = 0, min = 0;
	double  dis = 0;
	while(scanf("%d %lf",&n,&d) != EOF)
	{
		k = 0;
		memset(used, 0, sizeof(used));
		memset(head, -1, sizeof(head));
		for(i = 1; i<=n; i++)
		{
			scanf("%d %d",&coor[i][0], &coor[i][1]);
			dis = sqrt(pow(coor[i][0],2)+pow(coor[i][1],2));
			if(dis-7.5<=d)
				add(0, i, dis-7.5);
		}
		for(i = 1; i<=n; i++)//转化为图
			for(j = 1; j<=n; j++)
			{
				if(i == j)
					continue;
				dis = sqrt((coor[i][0]-coor[j][0])*(coor[i][0]-coor[j][0])+(coor[i][1]-coor[j][1])*(coor[i][1]-coor[j][1]));
				if(dis<=d)
					add(i, j, dis);
			}
		for(i = 1; i<=n; i++)//点到岸边的路径转化为图
		{
			min = 50-coor[i][0];
			if(min>50-coor[i][1])
				min = 50-coor[i][1];
			if(min>50+coor[i][0])
				min = 50+coor[i][0];
			if(min>50+coor[i][1])
				min = 50+coor[i][1];
			if(min<=d)
				add(i, n+1, min);
		}
		if(d>=42.50)
			printf("42.50 1\n");
		else
		{
			dijkstar();
			if(dist[n+1] == 1234567890)
				printf("can't be saved\n");
			else
				printf("%.2lf %d\n",dist[n+1], step[n+1]);
		}
	}
	return 0;
}