1003. Emergency (25)
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
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
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.
5 6 0 2 1 2 1 5 3 0 1 1 0 2 2 0 3 1 1 2 1 2 4 1 3 4 1Sample Output
2 4
#include <iostream>
#include <fstream>
#include <algorithm>
#include <vector>
#include <cstring>
using namespace std;
//此代碼使用前，需刪除下面兩行+後面的system("PAUSE")
ifstream fin("in.txt");
#define cin fin
const int CITYNUM = 500;
const int INF = 0x7fffffff;
int city[CITYNUM]; //記錄各個城市的團隊數
int road[CITYNUM][CITYNUM]={0};
bool visited[CITYNUM]={false};
int minLen[CITYNUM]={0}; //從源城市到達index城市的最短路徑值
int sum[CITYNUM]={0}; //從源城市到達index城市，所能召集的最大團隊數
int same[CITYNUM]={0}; //從源城市到達index城市,具有相同最短的路徑個數
void Dij(int source,int dest,int n){ //dijkstra算法
int i,t,mm,next;
int count = 0;
int cur = source;
sum[cur]=city[cur];
same[cur]=1;
while(count< n-1){
visited[cur]=true;
mm=INF;
for(i=0;i<n;i++){
if(visited[i])continue;
if(road[cur][i]){
t = minLen[cur] + road[cur][i];
if(t < minLen[i] || minLen[i]==0){ //到達城市i，出現新的最短路徑
minLen[i]=t;
same[i]=same[cur]; //重新計數,可能到達本節點cur的最短路徑有多條
sum[i]=sum[cur]+city[i];
}else if(t == minLen[i]){ //到達城市i，出現相同的最短路徑
same[i]+=same[cur];
if(sum[cur]+city[i] > sum[i]) //記下團隊數較大的值
sum[i]=sum[cur]+city[i];
}
}
if(minLen[i] < mm && minLen[i]!=0){
mm = minLen[i];
next = i;
}
}
minLen[cur] = mm;
if(next == dest)break;
cur = next;
count++;
}
return;
}
int main()
{
int n,m,sc,dc;
cin>>n>>m>>sc>>dc;
int i;
for(i=0;i<n;i++)cin>>city[i];
int c1,c2;
for(i=0;i<m;i++){
cin>>c1>>c2;
cin>>road[c1][c2];
road[c2][c1]=road[c1][c2];
}
if(sc==dc){ //若所在地就是目的地 則直接輸出結果
cout<<1<<' '<<city[sc]<<endl;
return 0;
}
Dij(sc,dc,n);
cout<<same[dc]<<' '<<sum[dc]<<endl;
system("PAUSE");
return 0;
} PS：就是利用迪傑斯特拉來求最短路徑個數與路徑上的最大點權值之和。
利用矩陣存儲圖。
1.如果輸入的目的和源點相同，那麼直接輸出
2.使用same數據來記錄訪問到每個點，有幾條相同的路徑了。並且具有累加作用。如果不同，那麼就是前一個點的值，如果相同，那麼就是cur加上當前邊已有的。！！！訪問到每個未被訪問的點，都會遍歷它所在的road那一行。
3.求出距離最小的作爲下一個cur。當next是目的時，退出！。