UVA1001 Say Cheese（Dijkstra或Floyd）

题目链接：UVA1001

题意：在一个巨大奶酪中的A要以最短的时间与B相遇。在奶酪中走一米的距离花费的时间是10s，而奶酪中有许多洞，穿过这些洞的时间是0s。给出A、B以及各个洞的坐标，求最短的时间。

三维？？乖乖，这怎么用最短路算法。在搜了题解后才知道可以编号压缩成二维啊，这操作骚气，实在想不出来啊！！

思路：将起点，终点，各个洞进行编号看成一个一个的点，写一个函数求出各个点之间的距离（即边的权值），在运用dijstra或Floyd算法就可以了。Ps：求距离的时候可以将各个点看成一个一个的球，距离就是两球心之间的距离减去两个球的半径和。

数据类型要用double，WA到吐得节奏。

Floyd方法：

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <string>
#include <queue>
#include <map>
#define INF 0x3f3f3f3f
#define FRE() freopen("in.txt","r",stdin)

using namespace std;
typedef long long ll;
const int maxn = ;
int n;
double d[maxn][maxn];
struct H
{
    double x;
    double y;
    double z;
    double r;
}hole[maxn];

double dist(H &a,H &b)
{
    double x = (a.x-b.x)*(a.x-b.x);
    double y = (a.y-b.y)*(a.y-b.y);
    double z = (a.z-b.z)*(a.z-b.z);
    if(sqrt(x+y+z) - a.r - b.r > 0.0)
        return sqrt(x+y+z) - a.r - b.r;
    else
        return 0.0;
}

int main()
{
   // FRE();
    int cnt = ;
    while(scanf("%d",&n) && n != -)
    {
        for(int i = ; i < n; i++)
        {
            scanf("%lf%lf%lf%lf",&hole[i].x,&hole[i].y,&hole[i].z,&hole[i].r);
        }
        scanf("%lf%lf%lf",&hole[n].x,&hole[n].y,&hole[n].z);
        scanf("%lf%lf%lf",&hole[n+].x,&hole[n+].y,&hole[n+].z);

        hole[n].r = hole[n+].r = ;

        for(int i = ; i <= n+; i++)
        for(int j = ; j <= n+; j++)
        {
            if(i == j)
                d[i][j] = ;
            else
                d[i][j] = INF;
        }
        for(int i = ; i < n+; i++)
        for(int j = ; j < n+; j++)
            if(i != j) d[i][j] = dist(hole[i],hole[j]);

        for(int k = ; k < n+; k++)
        for(int i = ; i < n+; i++)
        for(int j = ; j < n+; j++)
        {
            d[i][j] = min(d[i][j], d[i][k]+d[k][j]);
        }
        d[n][n+] *= ;
        printf("Cheese %d: Travel time = %.0f sec\n",++cnt,d[n][n+]);
    }
    return ;
}

Dijkstra邻接矩阵方法：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>
#define FRE() freopen("in.txt","r",stdin)
#define INF 0x3f3f3f3f

using namespace std;
const int maxn = ;
double x[maxn],y[maxn],z[maxn],r[maxn];
double d[maxn],vis[maxn];
double mp[maxn][maxn];
int n;

double dist(int i,int j)
{
    double tx = (x[i]-x[j])*(x[i]-x[j]);
    double ty = (y[i]-y[j])*(y[i]-y[j]);
    double tz = (z[i]-z[j])*(z[i]-z[j]);
    double res = sqrt(tx + ty + tz) - r[i] - r[j];
    if(res > )
        return res;
    else
        return ;
}

void Dij()
{
    memset(vis,,sizeof(vis));
    for(int i = ; i <= n+; i++)
        d[i] = INF;
    d[] = ;
    for(int i = ; i <= n+; i++)
    {
        int u;double mmin = INF;
        for(int i = ; i <= n+; i++)
        {
            if(!vis[i] && d[i] < mmin)
            {
                mmin = d[i];
                u = i;
            }
        }
        if(u == n+) return;
        vis[u] = ;
        for(int i = ; i <= n+; i++)
        {
            d[i] = min(d[i], d[u]+mp[u][i]);
        }
    }

}

int main()
{
    //FRE();
    int cnt = ;
    while(scanf("%d",&n) && n != -)
    {
        for(int i = ; i <= n; i++)
        {
            scanf("%lf%lf%lf%lf",&x[i],&y[i],&z[i],&r[i]);
        }
        scanf("%lf%lf%lf",&x[],&y[],&z[]);r[] = ;
        scanf("%lf%lf%lf",&x[n+],&y[n+],&z[n+]);r[n+] = ;

        for(int i = ; i <= n+; i++)
        {
            for(int j = ; j <= n+; j++)
                mp[i][j] = dist(i,j);
        }
        Dij();
        printf("Cheese %d: Travel time = %.0f sec\n",++cnt,d[n+]*);

    }
    return ;
}

Dijkstra优先队列方法：vector数组的清空啊，别问我是怎么知道的！！！！！！

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>
#define FRE() freopen("in.txt","r",stdin)
#define INF 0x3f3f3f3f

using namespace std;
typedef pair<double,int> P;
const int maxn = ;
struct H
{
    double x,y,z;
    double r;
}hole[maxn];
struct edge
{
    int to;
    double cost;
    edge(int t,double c):to(t),cost(c){}
};
vector<edge> g[maxn];
double d[maxn];

double dist(H &a,H &b)
{
    double x = (a.x - b.x)*(a.x - b.x);
    double y = (a.y - b.y)*(a.y - b.y);
    double z = (a.z - b.z)*(a.z - b.z);
    double res = sqrt(x + y + z) - a.r - b.r;
    if(res > )
        return res;
    else
        return ;
}

int main()
{
    //FRE();
    int cnt = ;
    int n;
    while(scanf("%d",&n) && n != -)
    {
        for(int i = ; i <= n; i++)
        {
            scanf("%lf%lf%lf%lf",&hole[i].x,&hole[i].y,&hole[i].z,&hole[i].r);
        }
        scanf("%lf%lf%lf",&hole[].x,&hole[].y,&hole[].z); hole[].r = ;
        scanf("%lf%lf%lf",&hole[n+].x,&hole[n+].y,&hole[n+].z);hole[n+].r = ;
        for(int i = ; i < n+; i++) g[i].clear();
        for(int i = ; i < n+; i++)
        {
            for(int j = i+; j < n+; j++)
            {
                double t = dist(hole[i],hole[j]);
                g[i].push_back(edge(j,t));
                g[j].push_back(edge(i,t));
            }
        }

        for(int i = ; i <n+; i++)
            d[i] = INF;
        d[] = ;
        priority_queue<P, vector<P>, greater<P> > que;
        que.push(P(,));
        while(!que.empty())
        {
            P p = que.top();
            que.pop();
            int v = p.second;
            if(d[v] < p.first) continue;
            for(int i = ; i < g[v].size(); i++)
            {
                edge ee = g[v][i];
                if(d[ee.to] > d[v] + ee.cost)
                {
                    d[ee.to] = d[v] + ee.cost;
                    que.push(P(d[ee.to], ee.to));
                }
            }
        }
        printf("Cheese %d: Travel time = %.0f sec\n",++cnt,d[n+]*);
    }
    return ;
}