A1003 Emergency

PAT (Advanced Level) Practice

1003 Emergency

分数 25

作者 CHEN, Yue

单位浙江大学

As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the length of each road between any pair of cities are marked on the map. When there is an emergency call to you from some other city, your job is to lead your men to the place as quickly as possible, and at the mean time, call up as many hands on the way as possible.

Input Specification:

Each input file contains one test case. For each test case, the first line contains 4 positive integers: N (≤500) - the number of cities (and the cities are numbered from 0 to N−1), M - the number of roads, C1 and C2 - the cities that you are currently in and that you must save, respectively. The next line contains N integers, where the i-th integer is the number of rescue teams in the i-th city. Then M lines follow, each describes a road with three integers c1, c2 and L, which are the pair of cities connected by a road and the length of that road, respectively. It is guaranteed that there exists at least one path from C1 to C2.

Output Specification:

For each test case, print in one line two numbers: the number of different shortest paths between C1 and C2, and the maximum amount of rescue teams you can possibly gather. All the numbers in a line must be separated by exactly one space, and there is no extra space allowed at the end of a line.

Sample Input:

Sample Output:

2 4

solution one：Dijkstra算法

#include <iostream>
#include <cstring>
#include <algorithm>using namespace std;const int INF = 0x3fffffff; //无穷大数据
const int maxn=510; //顶点最大值int G[maxn][maxn]; //邻接矩阵
int weight[maxn];  //顶点权值int w[maxn];  // 路径上顶点上筹集的最大物资
int num[maxn]; // 最短路径条数
int d[maxn];  //最短路径
bool hs[maxn]; // 顶点是否被标记int Nv,Ne,st,ed; //顶点数，边数，起点，终点void Read(); //读出数据，并建图
void Dijkstra(int s); //Gijkstra 算法int main()
{fill(*G,*G+maxn*maxn,INF);   //这句话没写，害我调试了半天Read();Dijkstra(st);cout << num[ed] << ' ' << w[ed] << endl;return 0;
}void Read()  //读出数据，并建图
{cin >> Nv >>Ne >> st >> ed;for(int i=0;i<Nv;++i)cin >> weight[i];for(int i=0;i<Ne;++i){int v1,v2,dis;cin >> v1 >> v2 >> dis;G[v1][v2] = G[v2][v1] = dis;}
}void Dijkstra(int s)  //Gijkstra 算法
{fill(d,d+maxn,INF);d[s]=0;w[s]=weight[s];num[s]=1;for(int i=0;i<Nv;++i){int u=-1,MIN=INF;for(int j=0;j<Nv;++j){if(hs[j]==0 && d[j]<MIN){u=j;MIN=d[j];}}if(u==-1)return;hs[u]=1;for(int v=0;v<Nv;++v){if(hs[v]==0 && G[u][v]!=INF && d[u]+G[u][v] < d[v]){d[v]= d[u] +G[u][v];w[v]=w[u]+weight[v];num[v]=num[u];}else if(hs[v]==0 && G[u][v]!=INF && d[u]+G[u][v] == d[v]){if(w[u]+weight[v] > w[v])w[v] = w[u]+weight[v];num[v]+=num[u];}}}
}

solution two:Bellman_Ford算法：

#include <iostream>
#include <cstring>
#include <algorithm>
#include <set>
#include <vector>using namespace std;typedef pair<int, int> PII;const int INF = 0x3fffffff; //无穷大数据
const int maxn=510; //顶点最大值vector<PII> Adj[maxn]; //邻接表
int weight[maxn];  //顶点权值int w[maxn];  // 路径上顶点上筹集的最大物资
int num[maxn]; // 最短路径条数
int d[maxn];  //最短路径
bool hs[maxn]; // 顶点是否被标记
set<int> pre[maxn];int Nv,Ne,st,ed; //顶点数，边数，起点，终点void Read(); //读出数据，并建图
void Dijkstra(int s); //Gijkstra 算法int main()
{Read();Dijkstra(st);cout << num[ed] << ' ' << w[ed] << endl;return 0;
}void Read()  //读出数据，并建图
{cin >> Nv >>Ne >> st >> ed;for(int i=0;i<Nv;++i)cin >> weight[i];for(int i=0;i<Ne;++i){int v1,v2,dis;cin >> v1 >> v2 >> dis;Adj[v1].push_back({v2,dis});Adj[v2].push_back({v1,dis});}
}void Dijkstra(int s)  //Gijkstra 算法
{fill(d,d+maxn,INF);d[s]=0;w[s]=weight[s];num[s]=1;for(int i=1;i<=Nv;++i){for(int u=0;u<Nv;++u){for(int j=0;j<Adj[u].size();++j){int v = Adj[u][j].first,dis = Adj[u][j].second;if(d[u] + dis < d[v]){d[v] = d[u] + dis;num[v] = num[u];w[v] = w[u] + weight[v];pre[v].clear();pre[v].insert(u);}else if(d[u] + dis == d[v]){if(w[u] + weight[v] > w[v])w[v] = w[u] + weight[v];pre[v].insert(u);num[v] = 0;for(auto s:pre[v])num[v] += num[s];}}}}
}

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