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;
//此代码使用前，需删除下面两行+后面的system("PAUSE")
ifstream fin("in.txt");
#define cin fin
const int CITYNUM = 500;
const int INF = 0x7fffffff;
int city[CITYNUM]; //记录各个城市的团队数
int road[CITYNUM][CITYNUM]={0};
bool visited[CITYNUM]={false};
int minLen[CITYNUM]={0}; //从源城市到达index城市的最短路径值
int sum[CITYNUM]={0}; //从源城市到达index城市，所能召集的最大团队数
int same[CITYNUM]={0}; //从源城市到达index城市,具有相同最短的路径个数
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){ //到达城市i，出现新的最短路径
minLen[i]=t;
same[i]=same[cur]; //重新计数,可能到达本节点cur的最短路径有多条
sum[i]=sum[cur]+city[i];
}else if(t == minLen[i]){ //到达城市i，出现相同的最短路径
same[i]+=same[cur];
if(sum[cur]+city[i] > sum[i]) //记下团队数较大的值
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 则直接输出结果
cout<<1<<' '<<city[sc]<<endl;
return 0;
}
Dij(sc,dc,n);
cout<<same[dc]<<' '<<sum[dc]<<endl;
system("PAUSE");
return 0;
}
PS：就是利用迪杰斯特拉来求最短路径个数与路径上的最大点权值之和。
利用矩阵存储图。
1.如果输入的目的和源点相同，那么直接输出
2.使用same数据来记录访问到每个点，有几条相同的路径了。并且具有累加作用。如果不同，那么就是前一个点的值，如果相同，那么就是cur加上当前边已有的。！！！访问到每个未被访问的点，都会遍历它所在的road那一行。
3.求出距离最小的作为下一个cur。当next是目的时，退出！。