hdu 2482

来源：互联网发布：如何获取车流量数据编辑：程序博客网时间：2024/06/17 07:32

神技能缩点

题意：
给出一幅地图起点和终点（要按题中所说的规则转化成实际坐标）
给出一些公共汽车站以及公车行进的线路
如果起点和终点距离不超过2000 则输出walk there
如果在离起点1000米内有车站离终点1000米内有车站
并且可以从起点的车站坐bus到终点的车站则输出最少换乘次数
否则输出take a taxi

#include <iostream>#include <cstdio>#include <cstring>#include <map>#include <vector>using namespace std;#define N 5010int ans[110][110];struct node{    int x,y;}data[N];void dis(char s[],int &x,int &y) {    double ans = 10240/2.0,a = 0,b = 0,bili = ans/10;    for(int i = 0; i < 8; i++) {        if(s[i] == '0') i = i;        else if(s[i] == '1') b = ans+b;        else if(s[i] == '2') a = ans+a;        else a = a+ans,b = b+ans;        ans = ans/2.0;        bili = ans/10;    }    x = a+x*bili;    y = b+y*bili;}double dist(node a,node b) {    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);}void solve() {    int n,m,k;scanf("%d",&n);    char s[30],s1[30];    map<string,int>mp; n = n + 2;    for(int i = 2; i < n; i++) {        scanf("%s %d %d",s,&data[i].x,&data[i].y);        mp[s] = i;    }    scanf("%d",&m);    int x,y; vector<int>vc[110];    m = m + 2;    for(int i = 2; i < m; i++) {        scanf("%d",&x);        while(x--) {            scanf("%s",&s);            y = mp[s];            vc[i].push_back(y);        }    }    if(dist(data[0],data[1]) < 4000000) {        printf("walk there\n");        return ;    }    for(int i = 0; i < m; i++) for(int j = 0; j < m; j++) ans[i][j] = N;    for(int i = 2; i < m; i++)      for(int j = 2; j < m; j++)        if(i!=j) {            bool flag = 0;            for(int k = 0; k < vc[i].size(); k++)              for(int z = 0; z < vc[j].size(); z++)                if(vc[i][k] == vc[j][z])                   flag = 1;            if(flag) ans[i][j] = ans[j][i] = 1;        } // 除了端点之外的边    for(int i = 2; i < m; i++) {  // 起点构边        for(int j = 0; j < vc[i].size(); j++)          if(dist(data[0],data[vc[i][j]]) <= 1000000)            ans[0][i] = 1;    }    for(int i = 2; i < m; i++) {   // 终点构边        for(int j = 0; j < vc[i].size(); j++)          if(dist(data[1],data[vc[i][j]]) <= 1000000) {            ans[1][i] = ans[i][1] = 1;          }    }       for(int k = 0; k < m; k++)      for(int i = 0; i < m; i++)        for(int j = 0; j < m; j++)          ans[i][j] = min(ans[i][j],ans[i][k]+ans[k][j]);    if(ans[0][1] >= N) printf("take a taxi\n");    else printf("%d\n",ans[0][1]-1);}int main() {    int t;scanf("%d",&t);    while(t--) {        char s[20]; scanf("%s %d %d",&s,&data[0].x,&data[0].y);        dis(s,data[0].x,data[0].y);        scanf("%s %d%d",&s,&data[1].x,&data[1].y);        dis(s,data[1].x,data[1].y);        solve();    }    return 0;}

0 0