标签:namespace priority 代码 vector org submit 注意 lan node
题目链接:https://cn.vjudge.net/problem/POJ-2253
一只Forg需要从节点1走到节点n
现要找一条各个间隔最小的路径
问间隔最小是多少
用dijsktra就好
查找间隔最小的路径
#include <cstdio>
#include <vector>
#include <queue>
#include <cmath>
using namespace std;
const int maxn=200, INF=0x3f3f3f3f;
const double eps=1e-6;
typedef pair<double, int> Pair;
struct Node{
int x, y;
Node(int x=0, int y=0):x(x), y(y) {}
}node[maxn+5];
struct Edge{
int from, to;
Edge(int from=0, int to=0):from(from), to(to) {}
};
vector<Edge> edges;
vector<int> G[maxn+5];
double dist[maxn+5];
void addEdge(int from, int to){
edges.push_back(Edge(from, to));
G[from].push_back(edges.size()-1);
G[to].push_back(edges.size()-1);
}
inline bool equal(const double &a, const double &b){
return (a-b<=eps && b-a<=eps);
}
inline double Dis(const int &a, const int &b){
return sqrt((node[a].x-node[b].x)*(node[a].x-node[b].x)+
(node[a].y-node[b].y)*(node[a].y-node[b].y));
}
double dij(void){
for (int i=0; i<=maxn; i++) dist[i]=INF;
priority_queue<Pair, vector<Pair>, greater<Pair> > que;
que.push(Pair(0, 1)); dist[1]=0;
while (que.size()){
Pair x=que.top(); que.pop();
if (!equal(x.first, dist[x.second])) continue;
int from=x.second;
for (int i=0; i<G[from].size(); i++){
Edge &e=edges[G[from][i]];
int to=(e.to==from)?e.from:e.to;
double dis=Dis(to, from), mdis=max(dist[from], dis);
if (dist[to]<mdis || equal(dist[to], mdis)) continue;
dist[to]=mdis;
que.push(Pair(dist[to], to));
}
}return dist[2];
}
int main(void){
int n, cnt=1, x, y;
while (scanf("%d", &n)==1 && n){
for (int i=1; i<=n; i++){
scanf("%d%d", &x, &y);
node[i]=Node(x, y);
for (int j=1; j<i; j++) addEdge(i, j);
}
printf("Scenario #%d\nFrog Distance = %.3f\n\n", cnt++, dij());
}
return 0;
}
Time | Memory | Length | Lang | Submitted |
---|---|---|---|---|
16ms | 1476kB | 1878 | G++ | 2018-05-23 15:31:25 |
POJ-2253 Frogger dijsktra查找间隔最小的路径
标签:namespace priority 代码 vector org submit 注意 lan node
原文地址:https://www.cnblogs.com/tanglizi/p/9094184.html