PAT A1072 Gas Station(30 分)-------圖最短路徑---比較難點的題
總結:
1.這道題用了dijstra演算法,關鍵是開始對G1非數字的處理即Gi處理成i+n;我最後一個測試點開始沒過就是因為用s.size()判斷輸入為數字還是G2,但是其實資料n+m是大於99的
程式碼:
#include<iostream>
#include<algorithm>
#include<vector>
#include<string>
using namespace std;
int G[1100][1100];
int n, m, k, ds;
const int inf = 99999999;
int dis[1100][1100];
int getid(string s)
{
if (s[0] != 'G'){ int k = stoi(s); return k; }
else { string p = s.substr(1, s.size() - 1); return stoi(p) + n; }
}
int main()
{
cin >> n >> m >> k >> ds;
fill(G[0], G[0] + 1100 * 1100, inf);
fill(dis[0], dis[0] + 1100 * 1100, inf);
for (int i = 1; i <= k; i++)
{
string pp1, pp2;
int p1, p2, dist;
cin >> pp1 >> pp2;
scanf("%d", &dist);
p1 = getid(pp1); p2 = getid(pp2);
if (p1 == p2)continue;
if (dist<G[p1][p2])G[p1][p2] = dist;
if (dist<G[p2][p1])G[p2][p1] = dist;
}
for (int p = n + 1; p <= n + m; p++) //
{
dis[p][p] = 0; int vi[1100];
fill(vi, vi + 1100, 0);
for (int i = 1; i <= n + m; i++)
{
int u = -1; int minlen = inf;
for (int j = 1; j <= n + m; j++)
{
if (dis[p][j]<minlen&&vi[j] == 0)
{
minlen = dis[p][j]; u = j;
}
}
if (u == -1)break;
vi[u] = 1;
for (int k = 1; k <= n + m; k++)
{
if (dis[p][u] + G[u][k] < dis[p][k] && G[u][k] != inf) //最短路徑不止一條????
{
dis[p][k] = dis[p][u] + G[u][k];
}
}
}
}
int minsum = 9999999; int minlen = -1; bool flag = true; int totalindex = -1; int index = 99999;
for (int i = n + 1; i <= n + m; i++)
{
int totalsum = 0; int totallen = inf; flag = true; totalindex = i;
for (int j = 1; j <= n; j++)
{
if (dis[i][j] > ds){ flag = false; break; }
totalsum +=dis[i][j];
if (dis[i][j]<totallen)totallen = dis[i][j];
}
if (flag == false)continue;
if (totallen > minlen){ minlen = totallen; minsum = totalsum; index = totalindex; }
else if (totallen == minlen&&totalsum<minsum){ minlen = totallen; minsum = totalsum; index = totalindex; }
else if (totallen == minlen&&totalsum == minsum&&totalindex<index){ minlen = totallen; minsum = totalsum; index = totalindex; }
}
if (flag == false)cout << "No Solution";
else {
double h1 = 1.0*minlen; double h2 = (1.0*minsum) / n;
cout << "G" << index - n << endl;
printf("%.1f %.1f", h1, h2);
}
return 0;
}