1003. Emergency (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 21 2 1 5 30 1 10 2 20 3 11 2 12 4 13 4 1

Sample Output

2 4

这道题目是典型的dfs问题，通过遍历临界矩阵，找到问题的解。

代码：

#include<iostream>#include<cstdlib>#include<vector>#include<cstring>using namespace std;int n,m,c1,c2;int **g;int *flag;int *team;vector<int> path;//存放路径（此问题虽然用不到，但是有些深搜问题需要输出最短路径）int min_length=10000000;//存放最短路径int max_team;//存放最多队伍int pl;//临时存放路径长度int mt;//临时存放最大队伍数量int shortnum=1;//存放最短路径的条数void dfs(int s);int main(){cin>>n>>m>>c1>>c2;g=(int **)malloc(n*sizeof(int *));flag=(int *)malloc(n*sizeof(int));memset(flag,0,4*n);for(int i=0;i<n;i++){g[i]=(int *)malloc(n*sizeof(int));memset(g[i],0,4*n);}team=(int *)malloc(n*sizeof(int));for(int i=0;i<n;i++){cin>>team[i];}for(int i=0;i<m;i++){int a,b,c;cin>>a>>b>>c;g[a][b]=c;g[b][a]=c;}flag[c1]=1;path.push_back(c1);mt+=team[c1];dfs(c1);cout<<shortnum<<" "<<max_team;return 0;}void dfs(int s){if(s==c2){//当搜索到目的地时int f=0;if(pl<min_length){shortnum=1;max_team=0;min_length=pl;f=1;}else if(pl==min_length){shortnum++;f=1;}if(mt>max_team&&f==1) max_team=mt;return;}for(int i=0;i<n;i++){   if(g[s][i]==0) continue;   if(flag[i]!=0) continue;   flag[i]=1;   pl+=g[s][i];   mt+=team[i];   path.push_back(i);   dfs(i);   path.pop_back();   flag[i]=0;   mt-=team[i];   pl-=g[s][i];}}