【题解】PATA-1003 Emergency

Emergency(PATA-1003)

题面

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.

输入

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.

输出

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.

样例输入

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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

样例输出

1
2 4

提示

思路

给你一个连通图,及其点权和边权。求起点c1到终点c2的最短路径数量,以及最短路径上点权最大和。

跑Dijkstra,在$dis[u]+w \le dis[v]$的时候,分类讨论,更新最短路径数量和点权最大和。

代码

查看代码
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const int inf = 0x3f3f3f3f;
const int mxn = 1e3 + 5;

struct E {
int to, next, w;
}e[mxn];

int H[mxn], tot;

void add(int from, int to, int w){
e[tot] = {to, H[from], w};
H[from] = tot++;
}

void graph_init(int n)
{
for(int i=0; i<n; i++)
H[i] = -1;
tot = 0;
}

int dis[mxn], num[mxn], man[mxn];
bool vis[mxn];
priority_queue<pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > > q;

int t[mxn];

void dijkstra_init(int n)
{
for(int i=0; i<n; i++){
dis[i] = inf;
man[i] = vis[i] = 0;
}
}

void dijkstra(int s, int n)
{
num[s] = 1; man[s] = t[s];
q.push({dis[s]=0, s});
while(!q.empty()){
int u = q.top().second; q.pop();
if(vis[u]) continue;
vis[u] = 1;
for(int i=H[u]; ~i; i=e[i].next){
int v=e[i].to, w=e[i].w;
if(!vis[v] && dis[u]+w < dis[v]){
dis[v] = dis[u]+w;
q.push({dis[v], v});
num[v] = num[u];
man[v] = max(man[v], man[u]+t[v]);
}
else if(!vis[v] && dis[u]+w == dis[v]){
num[v] += num[u];
man[v] = max(man[v], man[u]+t[v]);
}
}
}
}

int main()
{
int n, m, c1, c2;
scanf("%d %d %d %d", &n, &m, &c1, &c2);

for(int i=0; i<n; i++)
scanf("%d", &t[i]);

graph_init(n);
for(int i=0; i<m; i++){
int u, v, w;
scanf("%d %d %d", &u, &v, &w);
add(u, v, w);
add(v, u, w);
}

dijkstra_init(n);
dijkstra(c1, n);

printf("%d %d\n", num[c2], man[c2]);
return 0;
}
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