1003 Emergency (25分)
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 (≤) - 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 Specification:
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
题意：计算从救援队当前所在的城市到目标城市的通路数，途中经过的城市会有救援人员，求不同的通路上救援人员的最大值。
思路：（边）最短路问题 + （顶点）最大价值
AC 代码：
#include<iostream>
#include<algorithm>
using namespace std;
int n, m, c1, c2;
int e[510][510], weight[510], dis[510], num[510], w[510];
bool visit[510];
const int inf = 99999999;
int main() {
scanf("%d%d%d%d",&n, &m, &c1, &c2);
for(int i = 0; i < n; ++i) {
scanf("%d", &weight[i]);
}
fill(e[0], e[0] + 510 * 510, inf);
fill(dis, dis + 510, inf);
int a, b, c;
for (int i = 0; i < m; ++i) {
scanf("%d%d%d", &a, &b, &c);
e[a][b] = e[b][a] = c;
}
dis[c1] = 0;
w[c1] = weight[c1];
num[c1] = 1;
for (int i = 0; i < n; ++i) {
int u = -1, minn = inf;
for (int j = 0; j < n; ++j) {
if (visit[j] == false && dis[j] < minn) {
u = j;
minn = dis[j];
}
}
if (u == -1) break;
visit[u] = true;
for (int v = 0; v < n; ++v) {
if (dis[u] + e[u][v] < dis[v]) {
dis[v] = dis[u] + e[u][v];
num[v] = num[u];
w[v] = w[u] + weight[v];
} else if (dis[u] + e[u][v] == dis[v]) {
num[v] = num[v] + num[u];
if (w[u] + weight[v] > w[v])
w[v] = w[u] + weight[v];
}
}
}
printf("%d %d", num[c2], w[c2]);
return 0;
}
2021-01-29
用DFS来做
python版：
# 信息读入 num_city, num_roads, cur_city, save_city = list(map(int, input().split())) rescue = list(map(int, input().split())) roads = [[] for _ in range(num_city)] for _ in range(num_roads): a = list(map(int, input().split())) roads[a[0]].append((a[1], a[2])) roads[a[1]].append((a[0], a[2])) # 定义几个变量 min_roads, max_rescue, min_distance = 0, 0, 99999 temp_distance, temp_rescue = 0, rescue[cur_city] visited = {cur_city} def dfs(city): global save_city, temp_distance, min_distance, temp_rescue, min_roads, max_rescue, visited if city == save_city: if temp_distance < min_distance: min_distance = temp_distance min_roads = 1 max_rescue = temp_rescue elif temp_distance == min_distance: min_roads += 1 if temp_rescue > max_rescue: max_rescue = temp_rescue return for next_city, distance in roads[city]: if next_city not in visited: visited.add(next_city) temp_distance += distance temp_rescue += rescue[next_city] dfs(next_city) temp_distance -= distance temp_rescue -= rescue[next_city] visited.remove(next_city) dfs(cur_city) print(min_roads, max_rescue)
参考：
https://www.liuchuo.net/archives/2359
https://qsctech-sange.github.io/1003-Emergency.html#python3
