FDNY to the Rescue!
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 2897 Accepted: 887
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 2897 Accepted: 887
Description
The Fire Department of New York (FDNY) has always been proud of their response time to fires in New York City, but they want to make their response time even better. To help them with their response time, they want to make sure that the dispatchers know the closest firehouse to any address in the city. You have been hired to write this software and are entrusted with maintaining the proud tradition of FDNY. Conceptually, the software will be given the address of the fire, the locations of the firehouses, street intersections, and the time it takes to cover the distance between each intersection. It will then use this information to calculate how long it takes to reach an address from each firehouse.
Given a specific fire location in the city, the software will calculate the time taken from all the fire stations located in the city to reach the fire location. The list of fire stations will be sorted from shortest time to longest time. The dispatcher can then pick the closest firestation with available firefighters and equipment to dispatch to the fire.
Input
Line 1:
# of intersections in the city, a single integer (henceforth referred to as N) N<20
Lines 2 to N+1:
A table (square matrix of integer values separated by one or more spaces) representing the time taken in minutes between every pair of intersections in the city. In the sample input shown below the value "3" on the 1st row and the 2nd column represents the time taken from intersection #1 to reach intersection #2.
Similarly the value "9" on the 4th row and the 2nd column represents the time taken from intersection #4 to reach intersection #2.
A value of -1 for time means that it is not possible to go directly from the origin intersection (row #) to the destination intersection (column #). All other values in the table are non-negative.
Line N+2:
An integer value n (<= N) indicating the intersection closest to the fire location followed by one or more integer values for the intersections closest to the fire stations (all on one line, separated by one or more spaces) will follow the input matrix.
Notes on input format:
1. The rows and columns are numbered from 1 to N.
2. All input values are integers
3. All fire locations are guaranteed reachable from all firehouses.
4. All distance calculations are made from the intersection closest to each firehouse to the intersection closest to the fire.
Output
Line 1:
A label line with the headings for each column, exactly as shown in the example.
Line 2 onwards (one line for each fire station):
A sorted list (based on time) showing the fire station (origin), the destination site, time taken and a complete shortest path of nodes from the originating fire station to the fire location.
Notes on output format:
1. Columns are tab separated.
2. If two or more firehouses are tied in time they can be printed in any order.
3. If more than one path exists that has the same minimal time for a given location & firehouse, either one can be printed on the output.
4. If the fire location and the fire station locations happen to be the same intersection, the output will indicate that the origin and destination have the same intersection number, the time will be "0" and the nodes in the shortest path will show just one number, the fire location.
Next is the picture for the sample input data.
Sample Input
6
0 3 4 -1 -1 -1
-1 0 4 5 -1 -1
2 3 0 -1 -1 2
8 9 5 0 1 -1
7 2 1 -1 0 -1
5 -1 4 5 4 0
2 4 5 6
In the above input the last line indicates that "2" is the location of the fire and "4", "5" and "6" are the intersections where fire stations are located.
Sample Output
Org Dest Time Path
5 2 2 5 2
4 2 3 4 5 2
6 2 6 6 5 2
题目大意：在一个地图上,找出从消防站到着火点的最短时间与最短时间中所走的点
只有一组数据
第一行代表一张地图,有n个点
接着是一个n×n的矩阵,代表从第i行j列中,i到j的时间,-1代表不能到达
最后一行,第一个数代表着火点,后面多个数代表消防站
输出的是消防站 着火点 时间 路径
解题思路：
这题要处理建图的关系，虽然图是无向图，但i到j的时间并不是j到i的时间，灭火大队又要从消防站出发到达火源，而火源是固定的，所以逆向建图是最好的处理，找从火源出发到达消防站的最短时间。就可以求出从所有点到着火点的最短路径，在算最短路径的时候记录路径点，这样从时间小的输出即可
#include<iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn = 25;
const int INF = 99999999;
int map[maxn][maxn],dis[maxn],lujing[maxn];
int n,fire,sum;//sum为消防站数量
int Is_station[maxn];
void dijkstra(){
int i,j,v,tmp;
bool vis[maxn];
memset(lujing,-1,sizeof(lujing));
memset(vis,false,sizeof(vis));
for(i=1;i<=n;i++) dis[i]=INF;
dis[fire]=0;
for(i=1;i<=n;i++){
tmp=INF;
for(j=1;j<=n;j++)
if(dis[j]<tmp && !vis[j]){
tmp=dis[j];
v=j;
}
vis[v]=true;
//这里是求火源到消防站的路径，在输出时，当lujing[j],j是消防站的位置，则倒着输出位置直到火源
for(j=1;j<=n;j++){
if(!vis[j] && map[v][j]!=-1 && dis[j]>dis[v]+map[v][j]){ //-1不能走
dis[j]=dis[v]+map[v][j];
lujing[j]=v; //记录路径
}
}
}
}
void output(){
int j,i,tmp,v;
printf("Org Dest Time Path
");
for(int j = 0;j<sum;j++){
tmp=INF;
for(i=1;i<=n;i++){ //每次找最小的时间输出
if(Is_station[i] && dis[i]<tmp){
tmp=dis[i];
v=i;
}
}
Is_station[v]=0; //将这个站点设为非消防站
printf("%d %d %d ",v,fire,dis[v]);
while(v!=-1){
printf("%d ",v);
v=lujing[v];
}
printf("
");
}
}
int main(){
int i,j,q;
scanf("%d",&n);
for(i=1;i<=n;i++){
for(j=1;j<=n;j++){
scanf("%d",&map[j][i]); //因为火源固定，边反向存储方便处理
}
}
scanf("%d",&fire);
sum=0;
memset(Is_station,0,sizeof(Is_station));
while(~scanf("%d",&q)) {
Is_station[q]=1; //标记消防站
sum++; //计算消防站总数
}
dijkstra();
output();
return 0;
}