PAT甲级 1003 Emergency （Dj最短路）
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.
Sample Input
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这题为什么不标记book[begin]想了好久。
原因就是这个最少的路径更新取决于前一个点 。到前一个点的最短路径有多少条，到这个点的最短路径就有多少条。
如果标记掉起点，则挨着起点的点无法更新路径条数（最少一条），会导致结果为零。
和dp有点类似，需要一层层往下更新，状态取决于前一个点。
代码：
#include
using namespace std;
int n,m,beg,des;
int c1,c2,l;
int temp,mmin;
int e[502][503],book[502],dis[502];
int num[502],w[502],total[502];
const int INF=99999999;
int main(){
cin>>n>>m>>beg>>des;
for(int i=0;i>w[i];
for(int i=0;i>c1>>c2>>l;
e[c1][c2]=e[c2][c1]=l;
}
for(int i=0;idis[temp]+e[temp][v]){
dis[v]=dis[temp]+e[temp][v];
num[v]=num[temp];
total[v]=total[temp]+w[v];
}
else if(dis[v]==dis[temp]+e[temp][v]){
num[v]+=num[temp];
if(total[temp]+w[v]>total[v])
total[v]=total[temp]+w[v];
}
}
}
}
cout<