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.
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.
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.
一道最短距离题，加上了两个标尺，有点权和最短路径的数量。
最短距离：d数组
最大点权：weight记录点权，w[u]记录s到u的最大点权和
最短路径数量：num[u]表示从s到u的最短路径数目
坑：输入的时候，因为把G[v][u] = G[u][v] 写成 G[u][v] = G[v][u] ，导致最后出现 0 0 ...还没赋值就又变成0...所以一定要细心！！！
AC
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn = 510; const int inf = 1000000000; int n, m, c1, c2; bool vis[maxn] = {false}; //已访问集合 int d[maxn], w[maxn]; //最短距离 最多救援队 int weight[maxn], num[maxn]; //点权值， 最短路径条数 int G[maxn][maxn]; void dij(int s) { fill(d, d+maxn, inf); memset(w, 0, sizeof(w)); memset(num, 0, sizeof(num)); d[s] = 0; w[s] = weight[s]; num[s] = 1; for(int i = 0; i < n; i++) { int u = -1, MIN = inf; for(int j = 0; j < n; j++) { if(vis[j] == false && d[j] < MIN) { u = j; MIN = d[j]; } } if(u == -1) return; vis[u] = true; for(int v = 0; v < n; v++) { if(G[u][v] != inf && vis[v] == false) { if(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(d[u] + G[u][v] == d[v]) { if(w[u] + weight[v] > w[v]) { w[v] = w[u] + weight[v]; } num[v] += num[u]; } } } } } int main() { scanf("%d%d%d%d", &n, &m, &c1, &c2); for(int i = 0; i < n; i++) { scanf("%d", &weight[i]); } int u,v; fill(G[0], G[0]+maxn*maxn,inf); for(int i = 0; i < m; i++) { scanf("%d%d", &u, &v); scanf("%d", &G[u][v]); G[v][u] = G[u][v]; } dij(c1); printf("%d %d\n", num[c2], w[c2]); return 0; }
