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 = 110; 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 = 0; while(scanf("%d",&n) && n != -1) { for(int i = 0; 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+1].x,&hole[n+1].y,&hole[n+1].z); hole[n].r = hole[n+1].r = 0; for(int i = 0; i <= n+1; i++) for(int j = 0; j <= n+1; j++) { if(i == j) d[i][j] = 0; else d[i][j] = INF; } for(int i = 0; i < n+2; i++) for(int j = 0; j < n+2; j++) if(i != j) d[i][j] = dist(hole[i],hole[j]); for(int k = 0; k < n+2; k++) for(int i = 0; i < n+2; i++) for(int j = 0; j < n+2; j++) { d[i][j] = min(d[i][j], d[i][k]+d[k][j]); } d[n][n+1] *= 10; printf("Cheese %d: Travel time = %.0f sec ",++cnt,d[n][n+1]); } return 0; }
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 = 300; 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 > 0) return res; else return 0; } void Dij() { memset(vis,0,sizeof(vis)); for(int i = 0; i <= n+1; i++) d[i] = INF; d[0] = 0; for(int i = 0; i <= n+1; i++) { int u;double mmin = INF; for(int i = 0; i <= n+1; i++) { if(!vis[i] && d[i] < mmin) { mmin = d[i]; u = i; } } if(u == n+1) return; vis[u] = 1; for(int i = 0; i <= n+1; i++) { d[i] = min(d[i], d[u]+mp[u][i]); } } } int main() { //FRE(); int cnt = 0; while(scanf("%d",&n) && n != -1) { for(int i = 1; i <= n; i++) { scanf("%lf%lf%lf%lf",&x[i],&y[i],&z[i],&r[i]); } scanf("%lf%lf%lf",&x[0],&y[0],&z[0]);r[0] = 0; scanf("%lf%lf%lf",&x[n+1],&y[n+1],&z[n+1]);r[n+1] = 0; for(int i = 0; i <= n+1; i++) { for(int j = 0; j <= n+1; j++) mp[i][j] = dist(i,j); } Dij(); printf("Cheese %d: Travel time = %.0f sec ",++cnt,d[n+1]*10); } return 0; }
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 = 300; 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 > 0) return res; else return 0; } int main() { //FRE(); int cnt = 0; int n; while(scanf("%d",&n) && n != -1) { for(int i = 1; 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[0].x,&hole[0].y,&hole[0].z); hole[0].r = 0; scanf("%lf%lf%lf",&hole[n+1].x,&hole[n+1].y,&hole[n+1].z);hole[n+1].r = 0; for(int i = 0; i < n+2; i++) g[i].clear(); for(int i = 0; i < n+2; i++) { for(int j = i+1; j < n+2; 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 = 0; i <n+2; i++) d[i] = INF; d[0] = 0; priority_queue<P, vector<P>, greater<P> > que; que.push(P(0,0)); while(!que.empty()) { P p = que.top(); que.pop(); int v = p.second; if(d[v] < p.first) continue; for(int i = 0; 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 ",++cnt,d[n+1]*10); } return 0; }