POJ 3268_Silver Cow Party

题意：

n个地方，标号1~n，每个地方都有一头牛，现在要他们都去往标号为x的地方，再从x返回，每条道路都是单向的，求所有牛走的来回的最短路中的最大值。

分析：

注意在求每头牛走到x时，挨个算肯定超时，可以在将道路反向处理，都变成从x出。之前用vector模拟邻接表超时，后来用链表和数组分别模拟了邻接表，好像数组模拟的更快一些。

代码:

链表：

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
const int maxn =1005, maxm = 100005, INF = 0x3fffffff;
struct edge
{
   int to, w;
   edge* next;
};
int n, m, x;
int in[maxn], d[maxn], d1[maxn];
edge* head[maxn];
edge* head2[maxn];
queue<int>q;
void spfa()
{
    fill(d, d+1+n,INF);
    fill(in, in + 1 + n, 0);
    d[x] = 0;
    q.push(x);
    in[x] = 1;
    while(!q.empty()){
        int a = q.front();
        edge* t =head[a];q.pop();
        in[a] = 0;
        while(t!=NULL){
            if(d[t->to]>d[a]+t->w){
                d[t->to] = d[a] + t->w;
                if(!in[t->to]){q.push(t->to);in[t->to] = 1;}
            }
            t = t->next;
        }
    }
    return;
}
void spfa2()
{
    fill(d1, d1+1+n,INF);
    fill(in, in + 1 + n, 0);
    d1[x] = 0;
    q.push(x);
    in[x] = 1;
    while(!q.empty()){
        int a = q.front();
        edge* t=head2[a];q.pop();
        in[a] = 0;
        while(t!=NULL){
            if(d1[t->to]>d1[a]+t->w){
                d1[t->to] = d1[a] + t->w;
                if(!in[t->to]){q.push(t->to);in[t->to] = 1;}
            }
            t = t->next;
        }
    }
    return;
}
int main (void)
{
        scanf("%d%d%d",&n,&m,&x);
        int t1, t2, t3;
        int res = 0;
        memset(head, 0,sizeof(head));
        for(int i = 0; i <m; i++){
            scanf("%d%d%d", &t1, &t2, &t3);
            edge* temp=new edge;
            edge* temp2 = new edge;
            temp->to = t2;
            temp->w= t3;
            temp->next = NULL;
            if(head[t1]==NULL) head[t1] = temp;
            else {
                temp->next =head[t1];
                head[t1] = temp;
            }
            temp2->to = t1;
            temp2->w = t3;
            temp2->next = NULL;
            if(head2[t2]==NULL) head2[t2] = temp2;
            else {
                    temp2->next = head2[t2];
                    head2[t2] = temp2;
            }
        }
       spfa();
       spfa2();

       for(int i = 1; i <= n; i++){

        if(i!=x&&res<d[i]+d1[i]) res = d[i]+d1[i];
       }
       printf("%d
",res);
  }

数组：

#include<cstdio>
#include<iostream>
#include<queue>
using namespace std;
int n, m, x;
struct edge{
    int to,w;
    int next;
};
const int maxn =1005, maxm = 100005, INF = 0x3fffffff;
int head[maxn], head2[maxn];
edge e[maxm], re[maxm];
int in[maxn], d[maxn], d1[maxn];
void spfa()
{
    fill(d, d+n+1,INF);
    fill(in, in+n+1,0);
    queue<int>q;
    q.push(x);
    in[x] = 1;
    d[x] = 0;
    while(!q.empty()){
        int tmp = q.front();q.pop();
        in[tmp] = 0;
        int t = head[tmp];
        while(t!=-1){
            if(d[e[t].to] > d[tmp] + e[t].w){
                d[e[t].to] = d[tmp] + e[t].w;
                if(!in[e[t].to]){
                        q.push(e[t].to);
                        in[e[t].to] = 1;
            }
        }
         t = e[t].next;
    }
    }
}
void spfa2()
{
    fill(d1,d1+n+1,INF);
    fill(in, in+n+1,0);
    queue<int>q;
    q.push(x);
    in[x] = 1;
    d1[x] = 0;
    while(!q.empty()){
        int tmp = q.front();q.pop();
        in[tmp] = 0;
        int t = head2[tmp];
        while(t!=-1){
            if(d1[re[t].to] > d1[tmp]+re[t].w){
                d1[re[t].to] = d1[tmp] + re[t].w;
                if(!in[re[t].to]){
                        q.push(re[t].to);
                        in[re[t].to] = 1;
            }
        }
        t = re[t].next;
    }
    }
}
int main (void)
{
    scanf("%d%d%d",&n,&m,&x);
    int a, b ,c;
    fill(head, head+n+1, -1);
    fill(head2, head2+n+1, -1);
    for(int i = 0; i < m; i++){
            scanf("%d%d%d",&a,&b,&c);
            e[i].to = b, e[i].w = c;
            e[i].next = head[a];
            head[a] = i;
            re[i].to = a, re[i].w =c;
            re[i].next = head2[b];
            head2[b] = i;
    }
    spfa();
    spfa2();
    int res = 0;

    for(int i = 1; i <= n ; i++){
        if(i!=x&&res<d[i]+d1[i]) res = d[i]+d1[i];
    }
    printf("%d
",res);

}