uva 1001 Say Cheese 最短路 spfa Floyd

首先是建图，把下标为0和下标为n+1的两个点设为起点和终点，把这两个点看成两个半径为0的洞，下标1~n的点为洞，则图上两点的距离为dis = |(xi,yi,zi)-(xj,yj,zj)|-ri-rj；如果dis<0，则边权为0，如果dis>0，则边权为10*dis。

有两种算法，一种是spfa+隐式图搜索，ac用时0.000，一种是直接Floyd，ac用时0.010。

spfa+隐式图

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define inf 0x3f3f3f3f
using namespace std;

struct Hole{
	int x, y, z, r;
	Hole(int a=0, int b=0, int c=0, int d=0):x(a),y(b),z(c),r(d){}
};

Hole hole[105];
int inq[105];
double d[105];
int n;

double get_dis(int i, int j)
{
	return sqrt((hole[i].x-hole[j].x)*(hole[i].x-hole[j].x)+(hole[i].y-hole[j].y)*(hole[i].y-hole[j].y)+(hole[i].z-hole[j].z)*(hole[i].z-hole[j].z))-hole[i].r-hole[j].r;                           
}

void spfa()
{
	queue que;
	memset(inq, 0, sizeof(inq));
	for(int i = 0; i < n+3; i++)	d[i] = inf;
	d[0] = 0;
	inq[0] = 1;
	que.push(0);
	while(!que.empty())
	{
		int u = que.front();
		que.pop();
		inq[u] = 0;
		for(int i = 0; i <= n+1; i++)
		{
			double dis;
			dis = (get_dis(u, i)<=0 ? 0:10*get_dis(u,i));
			if(d[u] < inf && d[i] > d[u]+dis)
			{
				d[i] = d[u]+dis;
				if(!inq[i])
				{
					que.push(i);
					inq[i] = 1;
				}
			}
		}
	}
}

int main()
{
	//freopen("ztest.txt","r",stdin);
	int t = 1;
	while(scanf("%d", &n) && n != -1)
	{
		int a, b, c, dd;
		for(int i = 1; i <= n; i++)
		{
			scanf("%d%d%d%d", &a, &b, &c, &dd);
			hole[i] = Hole(a, b, c, dd);
		}
		scanf("%d%d%d", &a, &b, &c);
		hole[0] = Hole(a, b, c, 0);
		scanf("%d%d%d", &a, &b, &c);
		hole[n+1] = Hole(a, b, c, 0);
		spfa();
		double ans = d[n+1];
		printf("Cheese %d: Travel time = %.0lf sec\n",t++,ans);
	}
	return 0;
}
		

Floyd

#include 
#include
#include
#include
using namespace std;

const int INF = 1000000;

int n;
double d[105][105];
int x[105], y[105], z[105], r[105];

void Floyd()
{
    for (int k = 1; k <= n+2;k++)
    for (int i = 1; i <= n+2;i++)
    for (int j = 1; j <= n+2; j++)
        d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
}

int main()
{
    //freopen("D:\\txt.txt", "r", stdin);
    int kase = 0;
    while (cin>>n && n!=-1)
    {
        for (int i = 1; i <= n + 2;i++)
        for (int j = 1; j <= n + 2;j++)
        if (i == j)  d[i][j] = 0;
        else d[i][j] = INF;

        for (int i = 1; i <= n; i++)
            cin >> x[i] >> y[i] >> z[i] >> r[i];
        for (int i = n + 1; i <= n + 2; i++)
        {
            cin >> x[i] >> y[i] >> z[i];
            r[i] = 0;
        }

        for (int i = 1; i <= n + 2;i++)
        for (int j = 1; j <= n + 2; j++)
        {
            if (i == j)  continue;
            double dis = sqrt((x[j] - x[i])*(x[j] - x[i]) + (y[j] - y[i])*(y[j] - y[i]) + (z[j] - z[i])*(z[j] - z[i]));
            dis = dis - r[j] - r[i];
            if (dis < 0)  dis = 0;
            d[i][j] = dis;
        }
        Floyd();
        printf("Cheese %d: Travel time = %.f sec\n", ++kase, 10 * d[n+1][n+2]);
    }
}