uva 1001 建图+最短路

来源：互联网发布：淘宝店多少好评一个钻编辑：程序博客网时间：2024/05/21 16:56

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<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<vector>#include<queue>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<Edge> edges;    vector<int> 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<HeapNode> 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

0 0