zoukankan      html  css  js  c++  java
  • 最短路总结

    最近过的最短路题目稍微总结一下,顺便写一下模板,最短路算法众多有floyd、dij、bell-man、spfa,速度最快就是dij+优先队列或者dij+堆排序,spfa理论上很快o(ke)但实际并不一定不过spfa传说中有一个很NB用处就是处理带负权回路。

    邻接表VS邻接矩阵:根据写题经验,如果可以用矩阵那一定是首选,矩阵速度比表快而且题目出现多重边时矩阵很好解决。
            今天不小心又遇见正向表与最短路图(HDU2433),感觉很兴奋哈,通宵刷题的感觉很好玩!遗憾的是hdu1385还没过,那是一题最短路路径问题等写过再去更新吧,最短路写得已经有点久料急着总结一下,今天就是通宵都要总结出,因为已经一天拖一天。最短路写完之后马上需要马上开始看数轮与数学料,光刷水题提高很慢。
    最短路:
    拓扑排序:
    最短路模版(复杂度O(nlogn))dij+优先队列:
    HDU1690
    #include<stdio.h>
    #include<queue>
    #include<string.h>
    #define inf 10000000000000
    using namespace std;
    __int64 mp[101][101];
    __int64 d[101];
    bool hash[101];
    __int64 a[101];
    __int64 l1,l2,l3,l4,c1,c2,c3,c4;
    __int64 n,m;
    __int64 start,end;
    typedef struct node
    {            
        __int64 adj;
        __int64 w;
        struct node*next;
    }node,*pnode;
    typedef struct Heap
    {
        bool operator<(Heap T)const
        {
            return T.dis<dis;
        }
        __int64 x;
        __int64 dis;
    }Heap;
    priority_queue<Heap>Q;
    __int64 bfs()
    {
        __int64 i,j;
        for(i=0;i<=100;i++)
            hash[i]=0,d[i]=inf;
        Heap min,in;
        pnode p;
        while(!Q.empty())
            Q.pop();
        in.x=start;
        in.dis=0;
        d[start]=0;
        Q.push(in);
        while(!Q.empty())
        {
            min=Q.top();
            Q.pop();
            if(min.x==end)
                return min.dis;
            if(hash[min.x])
                continue;
            hash[min.x]=true;
            for(int i=1;i<=n;i++)
            {
                if(d[i]>mp[min.x][i]+d[min.x]&&!hash[i])
                {
                    in.x=i;
                    in.dis=mp[min.x][i]+d[min.x];
                    d[i]=in.dis;
                    Q.push(in);
                }
            }
            
        }
        return -1;
    }
    int judge(int dis)
    {
        if(dis>0&&dis<=l1)
            return c1;
        else
            if(dis>l1&&dis<=l2)
                return c2;
            else
                if(dis>l2&&dis<=l3)
                    return c3;
                else if(dis>l3&&dis<=l4)
                    return c4;
                else
                    return -1;
    }
    int main()
    {
        __int64 p=1,i,j;
        int t;
        scanf("%d",&t);
        while(t--)
        {
            scanf("%I64d%I64d%I64d%I64d%I64d%I64d%I64d%I64d",&l1,&l2,&l3,&l4,&c1,&c2,&c3,&c4);
            scanf("%I64d%I64d",&n,&m);
            for(i=0;i<=n;i++)
                for(j=0;j<=n;j++)
                    mp[i][j]=inf;
            for(i=1;i<=n;i++)
                scanf("%I64d",&a[i]);
            for(i=1;i<=n;i++)
            {
                for(j=1;j<=n;j++)
                {
                    __int64 l=a[i]-a[j];
                    if(l<0)
                        l=-l;
                    __int64 s=judge(l);
                    if(s<0) s=inf;
                    mp[i][j]=s;
                    mp[j][i]=s;
                }
            }
            printf("Case %I64d: ",p++);
            for(i=0;i<m;i++)
            {
                scanf("%I64d%I64d",&start,&end);
                bfs();
                if(d[end]==inf)
                    printf("Station %I64d and station %I64d are not attainable. ",start,end);
                else
                    printf("The minimum cost between station %I64d and station %I64d is %I64d. ",start,end,d[end]);
            }
        }
        return 0;
    }
    SPFA模版:
    HDU1874
    #define MAX 999999999
    #include<queue>
    #include<stdio.h>
    #include<string.h>
    using namespace std;
    int n,m,mp[1001][1001],d[1001];
    bool visit[201];
    void init()
    {
        int i,j;
        for(i=0;i<=n;i++)
            for(j=0;j<=n;j++)
                mp[i][j]=MAX;
            for(i=0;i<=n;i++)
                d[i]=MAX,visit[i]=0;
    }
    void spfa(int s)
    {
        queue<int> k;
        int i,j;
        k.push(s);
        d[s]=0;
        visit[s]=1;
        while(!k.empty())
        {
            int h=k.front();
            k.pop();
            visit[h]=0;
            for(i=0;i<n;i++)
            {
                if(d[i]-mp[h][i]>d[h])
                {
                    d[i]=mp[h][i]+d[h];
                    if(!visit[i])
                    {
                        k.push(i);
                        visit[i]=1;
                    }
                }
            }
        }
    }
    int main()
    {
        int i,j;
        while(scanf("%d%d",&n,&m)!=EOF)
        {
            init();
            int x,y,w;
            for(i=0;i<m;i++)
            {
                scanf("%d%d%d",&x,&y,&w);
                mp[x][y]=mp[y][x]=(w<mp[x][y]?w:mp[x][y]);
            }
            int s,t;
            scanf("%d%d",&s,&t);
            spfa(s);
            if(d[t]==MAX) puts("-1");
            else printf("%d ",d[t]);
        }
        return 0;
    }
    HDU3342(拓扑排序模版)
    #include<stdio.h>
    #include<string.h>
    #include<set>
    using namespace std;
    struct khp
    {
        int in;
        set<int> k;
    };
    khp y[101];
    bool final[101];
    int n,m;
    bool topu()
    {
        int t,s=0,i,j;
        bool judge=false;
        for(i=0;i<n;i++)
            if(y[i].in==0)
            {
                t=i;
                break;
            }
            
            if(i==n) return false;
            final[t]=true;
            while(s<n)
            {
                judge=false;
            for(set<int>::iterator it=y[t].k.begin();it!=y[t].k.end();it++)
                y[*it].in--;
            for(i=0;i<n;i++)
                if(y[i].in==0&&!final[i])
                {
                    t=i;
                    s++;
                    final[t]=true;
                    judge=true;
                    break;
                }
            if(!judge)
                break;
            }
            if(s!=n-1)
                return false;
            else return true;
    }
    int main()
    {
        int i,j,a,b;
        while(scanf("%d%d",&n,&m)!=EOF)
        {
            for(i=0;i<101;i++)
            {
                final[i]=false;
                y[i].in=0;
                y[i].k.clear();
            }
            if(!n&&!m) break;
            for(i=0;i<m;i++)
            {
                scanf("%d%d",&a,&b);
                if(!y[b].k.count(a))
                {
                    y[b].k.insert(a);
                    y[a].in++;
                }
            }
            bool judge=topu();
            if(!judge)
                puts("NO");
            else puts("YES");
        }
        return 0
    }
     
  • 相关阅读:
    Consul常用命令
    ECharts 避免变窄
    TP3.2 日期默认格式
    新订单提示效果
    php 按照字典序排序 微信卡券签名算法用到
    td宽度自适应 窄的地方自动收缩
    git 删除本地分支,删除远程分支
    分页Model
    chrome表单自动填充如何取消
    tp3.2 如何比较两个字段
  • 原文地址:https://www.cnblogs.com/calmwithdream/p/5369885.html
Copyright © 2011-2022 走看看