PAT A1003

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 1
Sample Output
2 4

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
#include "stdio.h"
#include "math.h"
#include "string.h"
#include "iostream"
//#include "stdlib.h"
#include "vector"
#include "set"
#include "map"
//#include "stack"
#include "queue"
#include "algorithm"
using namespace std;

const int MAXV = 510;//最大顶点数
const int INF = 1000000000;//无穷大
//n为顶点数,m为边数,st和ed分别起点和终点
//G为邻接矩阵,weight为点权
//d[]记录最短距离,w[]记录最大点权之和,num[]记录最短路径条数
int n, m, st, ed, G[MAXV][MAXV], weight[MAXV];
int d[MAXV], w[MAXV], num[MAXV];
bool vis[MAXV] = {false};//vis[i]==true表示顶点i已访问,初始均为false

void Dijstra(int s){//s为起点
fill(d, d + MAXV, INF);
memset(num, 0, sizeof(num));
memset(w, 0, sizeof(w));
d[s] = 0;//s到s的距离是0
w[s] = weight[s];
num[s] = 1;
for (int i = 0; i < n; i++) {//循环n次
int u = -1, MIN = INF;//u使d[u]最小,MIN存放该最小的d[u]
for (int j = 0; j < n; j++) {//找到未访问的顶点中d[]最小的
if (vis[j] == false && d[j] < MIN) {
u = j;
MIN = d[j];
}
}
//找不到小于INF的d[u],说明剩下的顶点和起点s不连通
if (u == -1) return;
vis[u] = true;//标记u为已访问
for (int v = 0; v < n; v++) {
//如果v未访问&&u能到达v&&以u为中介点可以使d[v]更优
if (vis[v] == false && G[u][v] != INF){
if (d[u] + G[u][v] < d[v]) {//以u为中介点时能令d[v]变小
d[v] = d[u] + G[u][v];//覆盖d[v]
w[v] = w[u] + weight[v];//覆盖w[v]
num[v] = num[u];//覆盖num[v]
}else if (d[u] + G[u][v] == d[v]){//找到一条相同长度的路径
if (w[u] + weight[v] > w[v]) {//以u为中介点时点权之和更大
w[v] = w[u] + weight[v];//w[v]继承自w[u]
}
//注意最短路径条数与点权无关,必须写在外面
num[v] += num[u];
}
}
}
}
}

int main(){
scanf("%d%d%d%d", &n, &m, &st, &ed);
for (int i = 0; i < n; i++) {
scanf("%d", &weight[i]);//读入点权
}
int u, v;
fill(G[0], G[0] + MAXV * MAXV, INF);//初始化图G
for (int i = 0; i < m; i ++){
scanf("%d%d", &u, &v);
scanf("%d", &G[u][v]);//读入边权
G[v][u] = G[u][v];
}
Dijstra(st);//dijstra算法入口
printf("%d %d\n", num[ed], w[ed]);//最短距离条数,最短路径中的最大点权

return 0;
}

此题在最短路径的基础上增加了点权和最短路径条数
一般来讲会有三种出题方式:
1、增加边权 2、增加点权 3、求最短路径条数
这三种都是在d[u] + G[u][v] == d[v]的条件下判断
其中前两种类似于
c[v] = c[u] + cost[u][v];
w[v] = w[u] + weight[v];
后一种增加num[],初始化时令num[s] = 1;其余均为0;
当在条件出现的时候,num[v] += num[u];