8.22最短路径讲解

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今天训练赛做了个最短路径的题，floyd超时，老师说一般给的数据小与100你用floyd，其实想过dij，但是当从各个点（源点）到目标点用dij很麻烦，dij的源点是不变的，这个边怎么班，

One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires T_i (1 ≤ T_i ≤ 100) units of time to traverse.

Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow's return route might be different from her original route to the party since roads are one-way.

Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?

Input

Line 1: Three space-separated integers, respectively: N, M, and X
Lines 2.. M+1: Line i+1 describes road i with three space-separated integers: A_i,B_i, and T_i. The described road runs from farm A_i to farm B_i, requiring T_i time units to traverse.

Output

Line 1: One integer: the maximum of time any one cow must walk.

Sample Input

4 8 21 2 41 3 21 4 72 1 12 3 53 1 23 4 44 2 3

Sample Output

Hint

Cow 4 proceeds directly to the party (3 units) and returns via farms 1 and 3 (7 units), for a total of 10 time units.

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<iomanip>
using namespace std;
#define inf 0x3f3f3f3f
int dis1[1005],dis2[1005],edge[1005][1005],v[1005];
int n,m,x;
void dij1()
{
    for(int i=1;i<=n;i++)
        dis1[i]=edge[x][i];//!!!!!!!!!!!!!!!
        memset(v,0,sizeof(v));
        dis1[x]=0;
        v[x]=1;
        //cout<<3<<endl;
        for(int i=1;i<=n-1;i++)
        {//cout<<4<<endl;
            int minn,next;
            next=0;
            minn=inf;
            for(int j=1;j<=n;j++)
            {
                if(!v[j]&&minn>dis1[j])
                {
                    minn=dis1[j];
                    next=j;
                }
            }//cout<<5<<endl;
            if(next==0) break;
            v[next]=1;//cout<<111<<endl;
            for(int j=1;j<=n;j++)
            {
                if(!v[j]&&dis1[j]>dis1[next]+edge[next][j])
                    dis1[j]=dis1[next]+edge[next][j];//!!!!!!!!!!!!!!!!!!!
            }
        }//cout<<6<<endl;
}
void dij2()
{
    for(int i=1;i<=n;i++)
        dis2[i]=edge[i][x];//!!!!!!!!!!!!!!
        memset(v,0,sizeof(v));
        dis2[x]=0;
        v[x]=1;//cout<<7<<endl;
        for(int i=1;i<=n-1;i++)
        {//cout<<8<<endl;
            int minn,next;
            next=0;
            minn=inf;
            for(int j=1;j<=n;j++)
            {
                if(!v[j]&&minn>dis2[j])
                {
                    minn=dis2[j];
                    next=j;
                }
            }
            if(next==0) break;
            v[next]=1;//cout<<11<<endl;
            for(int j=1;j<=n;j++)
            {
                if(!v[j]&&dis2[j]>dis2[next]+edge[j][next])//!!!!!!!!!!!!!!!!!!!!
                    dis2[j]=dis2[next]+edge[j][next];//!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
            }//cout<<9<<endl;
        }
        //cout<<10<<endl;
        //cout<<22<<endl;
}
int main()
{
    int i,j,a,b,t,maxn;
    scanf("%d%d%d",&n,&m,&x);
    memset(edge,inf,sizeof(edge));
    for(i=1;i<=m;i++)
    {
        scanf("%d%d%d",&a,&b,&t);
        edge[a][b]=t;
    }
    //cout<<1<<endl;
    dij1();//cout<<2<<endl;
    dij2();
    maxn=0;
    for(i=1;i<=n;i++)
    {
        if(maxn<dis1[i]+dis2[i])
            maxn=dis1[i]+dis2[i];
    }
    cout<<maxn<<endl;
}
好题，怎么自己就是想不到呢。。。。，

第一个dij1()就是平唱的求源点到其他点的最短路径

dij2()求各个点到终点的最短路径，源点在变化，加的时候换一下坐标，就可以将终点作为原点，才是得到dis2（i)就是i到终点的最短路径

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