hihocoder 1138 Islands Travel dijkstra+heap 难度:3

很久不用dijkstra,几乎连基本性质也忘了,结果这道题就是考察dijkstra空间复杂度是o(n)

这里总结四种算法

算法名称           时间复杂度       空间复杂度

dijkstra+heap  O(elog(e+n))   O(n)

bellman-ford    O(ne)             O(n)

spfa                O(ke)             O(n)

floyd-warshall   O(n^3)          O(n^2)

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<queue>

using namespace std;
const int maxn = 1e5 + 5;
const int maxm = 1e6 + 6;

int first[maxn],n;
struct edge
{
    int t,c,nxt;
} e[maxm];

void addedge(int f,int t,int c,int ind)
{
    e[ind].nxt = first[f];
    e[ind].t = t;
    e[ind].c = c;
    first[f] = ind;
}

struct pnt
{
    int x,y,id;
    pnt()
    {
        x = y = id = 0;
    }
    pnt(int _x,int _y,int _id)
    {
        x = _x;
        y = _y;
        id = _id;
    }
};
bool cmpx(pnt p1,pnt p2)
{
    if(p1.x!= p2.x)return p1.x < p2.x;
    return p1.y < p2.y;
}
bool cmpy(pnt p1,pnt p2)
{
    if(p1.y!= p2.y)return p1.y < p2.y;
    return p1.x < p2.x;
}
pnt a[maxn];
long long dis[maxn];
bool vis[maxn];
typedef pair<long long ,int> P;
priority_queue<P, vector <P>, greater<P> > que;
long long dijkstra()
{
    for(int i = 0; i < n; i++)dis[i] = 2e18;
    memset(vis,false,sizeof vis);
    while(!que.empty())que.pop();

    dis[0] = 0;
    vis[0] = true;
    for(int p = first[0]; p != -1; p = e[p].nxt)
    {
        int t = e[p].t;
        dis[t] = e[p].c;
        que.push(P(dis[t],t));
    }

    while(!que.empty())
    {
        int f = que.top().second;
        que.pop();
        if(f == n-1)break;
        if(vis[f])continue;
        vis[f] = true;

        for(int p = first[f]; p != -1; p = e[p].nxt)
        {
            int t = e[p].t;
            if(dis[t] > dis[f] + e[p].c){
                dis[t] = dis[f] + e[p].c;
                que.push(P(dis[t],t));
            }
        }
    }

    return dis[n - 1];
}
int main()
{
    while(scanf("%d",&n)==1)
    {
        memset(first, -1, sizeof first);
        for(int i = 0; i < n; i++)
        {
            scanf("%d%d",&a[i].x,&a[i].y);
            a[i].id = i;
        }

        sort(a,a + n,cmpx);
        for(int i = 0; i < n - 1; i++)
        {
            addedge(a[i].id,a[i + 1].id,
                    min(abs(a[i].x - a[i + 1].x),abs(a[i].y - a[i + 1].y)),2 * i);
            addedge(a[i + 1].id,a[i].id,
                    min(abs(a[i].x - a[i + 1].x),abs(a[i].y - a[i + 1].y)),2 * i + 1);
        }
        sort(a,a + n,cmpy);
        for(int i = 0; i < n - 1; i++)
        {
            addedge(a[i].id,a[i + 1].id,
                    min(abs(a[i].x - a[i + 1].x),abs(a[i].y - a[i + 1].y)),2 * i + 2 * n);
            addedge(a[i + 1].id,a[i].id,
                    min(abs(a[i].x - a[i + 1].x),abs(a[i].y - a[i + 1].y)),2 * i + 1 + 2 * n);
        }

        long long ans = dijkstra();
        printf("%lld\n",ans);
    }
    return 0;
}