uva 1001 建图+最短路

题意：

在一个三维的空间内求从起点到终点的最短时间花费。

其中n个洞，在洞内通过的时间花费是0，在洞外的时间花费是每单位10sec

思路：

水题

建立起点到每个洞的边，建立终点到每个洞的边，建立起点到终点的边 ，边权是到达所需的时间花费。求一次最短路即可。

code：

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

const int maxn = 105;

int n;
struct ball{
    double x, y, z, r;
    ball(){}
    ball(double x_, double y_, double z_, double r_){
        x = x_;
        y = y_;
        z = z_;
    }
}bb[maxn];
double sx,sy,sz,ex,ey,ez;

const int INF = 0x3f3f3f3f;
const int MAXNODE = 110;
struct Edge{
    int u, v;
    double dist;
    Edge() {}
    Edge(int u, int v, double dist) {
        this->u = u;
        this->v = v;
        this->dist = dist;
    }
};
struct HeapNode {
    int u;
    double d;
    HeapNode() {}
    HeapNode(double d, int u) {
        this->d = d;
        this->u = u;
    }
    bool operator < (const HeapNode& c) const {
        return d > c.d;
    }
};
struct Dijkstra {
    int n, m;
    vector edges;
    vector g[MAXNODE];
    bool done[MAXNODE];
    double d[MAXNODE];
    int p[MAXNODE];

    void init(int tot) {
        n = tot;
        for (int i = 0; i < n; i++)
            g[i].clear();
        edges.clear();
    }

    void add_Edge(int u, int v, double dist) {
        edges.push_back(Edge(u, v, dist));
        m = edges.size();
        g[u].push_back(m - 1);
    }

    void dijkstra(int s) {
        priority_queue Q;
        for (int i = 0; i < n; i++) d[i] = (double)INF;
        d[s] = 0;
        memset(done, false, sizeof(done));
        Q.push(HeapNode(0, s));
        while (!Q.empty()) {
            HeapNode x = Q.top(); Q.pop();
            int u = x.u;
            if (done[u]) continue;
            done[u] = true;
            for (int i = 0; i < g[u].size(); i++) {
                Edge& e = edges[g[u][i]];
                if (d[e.v] > d[u] + e.dist) {
                    d[e.v] = d[u] + e.dist;
                    p[e.v] = g[u][i];
                    Q.push(HeapNode(d[e.v], e.v));
                }
            }
        }
    }
}gao;




double dist(ball b1, ball b2){
    double tmp = (b1.x - b2.x)*(b1.x - b2.x) + (b1.y - b2.y) * (b1.y - b2.y) + (b1.z - b2.z) * (b1.z - b2.z);
    return sqrt(tmp);
}
void init(){
    double x,y,z,r;
    for(int i = 0; i < n; i++){
        scanf("%lf%lf%lf%lf",&x,&y,&z,&r);
        bb[i].x = x;
        bb[i].y = y;
        bb[i].z = z;
        bb[i].r = r;
    }
    scanf("%lf%lf%lf",&sx,&sy,&sz);
    scanf("%lf%lf%lf",&ex,&ey,&ez);
    gao.init(n+2);
}
void deal(){
    ball b1(sx,sy,sz,0);
    for(int i = 0; i < n; i++){
        double w = dist(b1, bb[i]);
        w = w - bb[i].r;
        if(w < 0) w = 0;
        gao.add_Edge(i, n, w*10);
        gao.add_Edge(n, i, w*10);
    }
    ball b2(ex,ey,ez,0);
    for(int i = 0; i < n; i++){
        double w = dist(b2, bb[i]);
        w = w - bb[i].r;
        if(w < 0) w = 0;
        gao.add_Edge(i, n+1, w*10);
        gao.add_Edge(n+1, i, w*10);
    }

    for(int i = 0; i < n; i++){
        for(int j = i+1; j < n; j++){
            double w = dist(bb[i], bb[j]);
            w = w - bb[i].r - bb[j].r;
            if(w < 0) w = 0;
            gao.add_Edge(i, j, w*10);
            gao.add_Edge(j, i, w*10);
        }
    }

    double w = dist(b1, b2);
    gao.add_Edge(n, n+1, w*10);
    gao.add_Edge(n+1, n, w*10);
}
void solve(){
    gao.dijkstra(n);
    printf("%d sec\n",(int)(gao.d[n+1]+0.5));
}
int main(){
    int cas = 0;
    while(scanf("%d",&n) != EOF){
        if(n == -1) break;
        init();
        deal();
        printf("Cheese %d: Travel time = ",++cas);
        solve();
    }
    return 0;
}

注意最后的结果是四舍五入后的整形，这个地方wa了一发T_T