hdu2874 lca问题的Tarjan算法内存险过

最新推荐文章于 2019-12-14 16:22:37 发布

原创最新推荐文章于 2019-12-14 16:22:37 发布 · 498 阅读

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#acm #算法 #图论 #LCA #Tarjan

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本文探讨了在战后城市重建中如何解决材料运输路径问题。通过构建无圈且可能不连通的城市间道路网络模型，使用特定算法寻找任意两城市间的最短路径或判断是否连通。

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Problem Description

After World War X, a lot of cities have been seriously damaged, and we need to rebuild those cities. However, some materials needed can only be produced in certain places. So we need to transport these materials from city to city. For most of roads had been totally destroyed during the war, there might be no path between two cities, no circle exists as well.
Now, your task comes. After giving you the condition of the roads, we want to know if there exists a path between any two cities. If the answer is yes, output the shortest path between them.

Input

Input consists of multiple problem instances.For each instance, first line contains three integers n, m and c, 2<=n<=10000, 0<=m<10000, 1<=c<=1000000. n represents the number of cities numbered from 1 to n. Following m lines, each line has three integers i, j and k, represent a road between city i and city j, with length k. Last c lines, two integers i, j each line, indicates a query of city i and city j.

Output

For each problem instance, one line for each query. If no path between two cities, output “Not connected”, otherwise output the length of the shortest path between them.

Sample Input

Sample Output

Not connected
6

换了个姿势还是上了HDU 据学长给我的图论题库来说 lca问题也写完了

这题几乎还是模板题只不过是多了个判断是否在同一棵树上用了点小技巧再Tarjan时把根节点也传上去如果得结果时根节点不同就跳过

另外这题灰常容易爆内存稍微改了一下用 C++水过去了然而G++还是爆………………

ac code

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdlib>

using namespace std;
const int maxn=10005;
const int inf=0x3f3f3f3f;

struct node
{
    int v,d;
    node* nxt;
};

node *head1[maxn],edge1[maxn*2];
node *head2[maxn],edge2[2000002];
int root[maxn],ans[1000002],dis[maxn];
int vis[maxn];
int n,lnum,qnum,idx1,idx2;

void init()
{
    for(int i=0;i<=n;i++)
    {
        head1[i]=head2[i]=0;
        vis[i]=0;
    }
    idx1=idx2=0;
}

int fin(int x)
{
    if(root[x]==x) return x;
    return root[x]=fin(root[x]);
}

void add(int u,int v,int d,node *head[],node edge[],int& idx)
{
    edge[idx].v=v;
    edge[idx].d=d;
    edge[idx].nxt=head[u];
    head[u]=edge+idx++;

    edge[idx].v=u;
    edge[idx].d=d;
    edge[idx].nxt=head[v];
    head[v]=edge+idx++;
}

void Tarjan(int u,int rot)
{
    root[u]=u;
    vis[u]=rot;
    for(node* p=head2[u];p;p=p->nxt)
        if(vis[p->v]==rot)
            ans[p->d]=dis[u]+dis[p->v]-2*dis[fin(p->v)];
    for(node* p=head1[u];p;p=p->nxt)
        if(!vis[p->v])
        {
            dis[p->v]=dis[u]+p->d;
            Tarjan(p->v,rot);
            root[p->v]=u;
        }
}

int main()
{
    int u,v,d;
    while(scanf("%d%d%d",&n,&lnum,&qnum)!=EOF)
    {
        init();
        for(int i=1;i<=lnum;i++)
        {
            scanf("%d %d %d",&u,&v,&d);
            add(u,v,d,head1,edge1,idx1);
        }
        for(int i=1;i<=qnum;i++)
        {
            scanf("%d %d",&u,&v);
            ans[i]=inf;
            add(u,v,i,head2,edge2,idx2);
        }
        for(int i=1;i<=n;i++)
            if(!vis[i]) 
            {
                dis[i]=0;
                Tarjan(i,i);
            }
        for(int i=1;i<=qnum;i++)
        {
            if(ans[i]==inf) puts("Not connected");
            else printf("%d\n",ans[i]);
        }
    }
    return 0;
}