POJ 2387 Til the Cows Come Home Dijkstra求最短路径

Til the Cows Come Home

Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible. 

Farmer John's field has N (2 <= N <= 1000) landmarks in it, uniquely numbered 1..N. Landmark 1 is the barn; the apple tree grove in which Bessie stands all day is landmark N. Cows travel in the field using T (1 <= T <= 2000) bidirectional cow-trails of various lengths between the landmarks. Bessie is not confident of her navigation ability, so she always stays on a trail from its start to its end once she starts it. 

Given the trails between the landmarks, determine the minimum distance Bessie must walk to get back to the barn. It is guaranteed that some such route exists.

Input

* Line 1: Two integers: T and N 

* Lines 2..T+1: Each line describes a trail as three space-separated integers. The first two integers are the landmarks between which the trail travels. The third integer is the length of the trail, range 1..100.

Output

* Line 1: A single integer, the minimum distance that Bessie must travel to get from landmark N to landmark 1.

Sample Input

5 5
1 2 20
2 3 30
3 4 20
4 5 20
1 5 100

Sample Output

90

Hint

INPUT DETAILS: 

There are five landmarks. 

OUTPUT DETAILS: 

Bessie can get home by following trails 4, 3, 2, and 1.

 

题意：给你一个图，途中每一条边都带有权值（路径长度），求从1到n的最短路径。

思路：单元最短路问题，首先想到Dijkstra。在这里把最基础的Dij O(n^2)与邻接表存图、堆优化的Dij O((n+m)logn)作个比较。

ps：史上最坑题！没有之一！首先是题目和样例的t、n顺序不同很容易误导爆RE。。再一个就是重  边  问  题！基础Dij会WA。。这题还有数据范围也...总之还是先把基础打好。

 

基础Dij：贪心思想

Dijkstra 思路是维护一个集合 s ，集合内的点是已经确定最短路的点，可以视为一个大整体

每次操作找出距离这个集合源点最近的点（dis最小）加入集合中，

并确定它的最短路为它的上家的最短路+该边权值，存在 dis 中（松弛操作：更新该点与集合外直接相连点的dis）

这里通过b标记点是否存在于集合，n-1次操作即可，dis最终结果为各点与单源点之间的最短路径

 

ps：Dij不能跑有负权边的图和最长路(负权边可用spfa但不能处理负环，最长路的贪心是错的，可将边权值*-1转化为spfa求最短路），

Dij可通过priority_queue堆优化，每次pop弹出的点相当于加入集合的点，这里省去了扫最近点的时间

 

#include<stdio.h>
#include<string.h>
#include<limits.h>

int a[1010][1010];
int dis[1010],b[1010];
int n;

int min(int x,int y)
{
    return x<y?x:y;
}

void dij(int k)
{
    int i,j,mind,minj;
    memset(b,0,sizeof(b));
    for(i=1;i<=n;i++){
        dis[i]=INT_MAX;
    }
    dis[k]=0;
    for(i=1;i<n;i++){
        mind=INT_MAX;
        for(j=1;j<=n;j++){
            if(!b[j]&&dis[j]<mind){
                mind=dis[j];
                minj=j;
            }
        }
        b[minj]=1;
        for(j=1;j<=n;j++){
            if(!b[j]&&a[minj][j]>=0){
                dis[j]=min(dis[j],dis[minj]+a[minj][j]);
            }
        }
    }
}

int main()
{
    int t,x,y,z;
    memset(a,-1,sizeof(a));
    scanf("%d%d",&t,&n);
    while(t--){
        scanf("%d%d%d",&x,&y,&z);
        if(a[x][y]!=-1){
            if(z<a[x][y]){
                a[x][y]=z;
                a[y][x]=z;
            }
        }
        else{
            a[x][y]=z;
            a[y][x]=z;
        }
    }
    dij(1);
    printf("%d
",dis[n]);
    return 0;
}

优化Dij：

#include<stdio.h>
#include<string.h>
#include<limits.h>
#include<vector>
#include<queue>
using namespace std;

struct Node{
    int v,w;
    friend bool operator<(Node a,Node b)
    {
        return a.w>b.w;
    }
}node;

vector<Node> a[1005];
int dis[1005];
int n;

void dij(int k)
{
    int v1,v2,i;
    priority_queue<Node> q;
    dis[k]=0;
    node.w=0;
    node.v=k;
    q.push(node);
    while(q.size()){
        v1=q.top().v;
        q.pop();
        for(i=0;i<a[v1].size();i++){
            v2=a[v1][i].v;
            if(dis[v2]>dis[v1]+a[v1][i].w){
                dis[v2]=dis[v1]+a[v1][i].w;
                node.w=dis[v2];
                node.v=v2;
                q.push(node);
            }
        }
    }
}

int main()
{
    int t,x,y,z,i;
    scanf("%d%d",&t,&n);
    for(i=1;i<=n;i++){
        a[i].clear();
        dis[i]=INT_MAX;
    }
    while(t--){
        scanf("%d%d%d",&x,&y,&z);
        node.w=z;
        node.v=y;
        a[x].push_back(node);
        node.v=x;
        a[y].push_back(node);
    }
    dij(1);
    printf("%d
",dis[n]);
    return 0;
}