題干：

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 Specification:

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,?C?1???and?C?2???- 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?c?1??,?c?2???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?C?1???to?C?2??.

Output Specification:

For each test case, print in one line two numbers: the number of different shortest paths between?C?1???and?C?2??, 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

題目大意：

作為一個(gè)城市的緊急救援隊(duì)隊(duì)長，你會得到一份特殊的國家地圖。這張地圖顯示了由一些道路連接起來的幾個(gè)分散的城市。地圖上標(biāo)明了每個(gè)城市的救援隊(duì)數(shù)量和每對城市之間的道路長度。當(dāng)從其他城市接到緊急電話時(shí)，你的工作就是盡快把你的人帶到那個(gè)地方，同時(shí)在路上盡可能多地召集人手。

一句話題意：給定一張無向圖，給定起點(diǎn)和終點(diǎn)，點(diǎn)有點(diǎn)權(quán)，邊有邊權(quán)，求兩點(diǎn)之間最短路條數(shù)，并求出在最短路的條件下路徑的點(diǎn)權(quán)和的最大值。

解題報(bào)告：

不得不說，，，幾年前的PAT考題確實(shí)比現(xiàn)在要難得多，這題雖然是裸題，但是代碼量也不小了，竟然才是25分題。

直接Dijkstra搞雙權(quán)值的就行，記得點(diǎn)編號是0~N-1的，所以要先轉(zhuǎn)化到1~N。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<stack> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define FF first #define SS second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 2e5 + 5; const int INF = 0x3f3f3f3f; int n,m,st,ed,val[MAX]; struct Edge {int v,ne,w; } e[MAX*2]; int head[MAX],tot; void add(int u,int v,int w) {e[++tot].v = v;e[tot].w = w;e[tot].ne = head[u];head[u] = tot; } struct Point {int pos,c;Point(){}Point(int pos,int c):pos(pos),c(c){}bool operator<(const Point & b) const {return c > b.c; } }; priority_queue<Point> pq; int dis[MAX],vis[MAX],ans[MAX],Max[MAX]; void Dij() {for(int i = 1; i<=n; i++) dis[i] = INF,vis[i] = 0,ans[i] = 0,Max[i] = 0;pq.push(Point(st,0)); dis[st] = 0; ans[st] = 1; Max[st] = val[st];while(pq.size()) {Point cur = pq.top();pq.pop();if(vis[cur.pos] == 1) continue;vis[cur.pos] = 1;for(int i = head[cur.pos]; ~i; i = e[i].ne) {int v = e[i].v;if(vis[v]) continue;if(dis[v] > dis[cur.pos] + e[i].w) {dis[v] = dis[cur.pos] + e[i].w;ans[v] = ans[cur.pos];Max[v] = Max[cur.pos] + val[v];pq.push(Point(v,dis[v]));}else if(dis[v] == dis[cur.pos] + e[i].w) {ans[v] += ans[cur.pos];Max[v] = max(Max[v],Max[cur.pos] + val[v]);}}} } int main() {cin>>n>>m>>st>>ed; st++,ed++; for(int i = 1; i<=n; i++) cin>>val[i],head[i] = -1;for(int u,v,w,i = 1; i<=m; i++) {scanf("%d%d%d",&u,&v,&w); u++,v++;add(u,v,w);add(v,u,w);}Dij();printf("%d %d\n",ans[ed],Max[ed]);return 0 ; }

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