poj 2394 Checking a Alibi-dijkstra

Checking an Alibi
Time Limit: 1000MS  Memory Limit: 65536K
Total Submissions: 5521  Accepted: 2009

Description

A crime has been comitted: a load of grain has been taken from the barn by

one of FJ's cows. FJ is trying to determine which of his C (1 <= C <= 100)

cows is the culprit. Fortunately, a passing satellite took an image of his

farm M (1 <= M <= 70000) seconds before the crime took place, giving the

location of all of the cows. He wants to know which cows had time to get

to the barn to steal the grain.

Farmer John's farm comprises F (1 <= F <= 500) fields numbered 1..F and

connected by P (1 <= P <= 1,000) bidirectional paths whose traversal time

is in the range 1..70000 seconds (cows walk very slowly). Field 1 contains

the barn. It takes no time to travel within a field (switch paths).

Given the layout of Farmer John's farm and the location of each cow when

the satellite flew over, determine set of cows who could be guilty.

NOTE: Do not declare a variable named exactly 'time'. This will reference

the system call and never give you the results you really want.
Input

* Line 1: Four space-separated integers: F, P, C, and M

* Lines 2..P+1: Three space-separated integers describing a path: F1,F2,

and T. The path connects F1 and F2 and requires T seconds to traverse.

* Lines P+2..P+C+1: One integer per line, the location of a cow. The first

line gives the field number of cow 1, the second of cow 2, etc.
Output

* Line 1: A single integer N, the number of cows that could be guilty of

the crime.

* Lines 2..N+1: A single cow number on each line that is one of the cows

that could be guilty of the crime. The list must be in ascending order.
Sample Input

7 6 5 8
1 4 2
1 2 1
2 3 6
3 5 5
5 4 6
1 7 9
1
4
5
3
7
Sample Output

4
1
2
3
4
Hint

INPUT DETAILS:

Fields/distances like this:

          6
      4------5
      |      |
     2|      |
      |      |
7-----1      |5
   9  |      |
     1|      |
      |      |
      2------3

OUTPUT DETAILS:

Any cow except cow 5 could have done it. Cow 5 would take 9 seconds to get

to the barn
//错了n次。代码虽然水，但是就是初始化的问题还是存在，调用init函数居然wa，汗
果断改成：memset(map,0x3f,sizeof(map));
讲一下题目的意思：求从顶点1到各个顶点满足路径小于M的结点的顺序，记录下个数并

输出各个结点的顺序

代码：

#include<cstdio>#include<string>#include<cstring>#include<cmath>#include<queue>#include<iostream>#include<algorithm>using namespace std;#define INF 0x3f3f3f3f#define MAX 505#define pf printfint map[MAX][MAX];int dis[MAX];  bool vis[MAX];  int ans[105];int F,P,C,M;void Dijkstra(){memset(vis,false,sizeof(vis));for(int i=1;i<=F;i++)dis[i]=INF;dis[1]=0;for (int i=1;i<=F;i++) {int k=0,MIN=INF;for (int j=1;j<=F;j++) {if (!vis[j]&& dis[j]<MIN){MIN=dis[j];k=j;}}vis[k]=true;for(int j=1;j<=F;j++) {if(map[k][j]!=INF &&dis[j]>dis[k]+map[k][j])dis[j]=dis[k]+map[k][j];}}}void init(){for(int i=1;i<=F;i++)for(int j=1;j<=F;j++){map[i][j]=0x3f;}}int main () {scanf("%d%d%d%d",&F,&P,&C,&M);int i,a,b,T;memset(map,0x3f,sizeof(map));//init();memset(ans,0,sizeof(ans));for(int i=0;i<P;i++){scanf("%d%d%d",&a,&b,&T);map[a][b]=map[b][a]=min(map[a][b],T);}Dijkstra();int cow,num=0;for(int i=1;i<=C;i++){scanf("%d",&cow);if (dis[cow]<=M)ans[num++]=i;}pf("%d\n",num);for(int i=0;i<num;i++)pf("%d\n",ans[i]);return 0;}