hdu 2433 Travel

Posted 2020-10-24 日拱一卒 功不唐捐

http://acm.hdu.edu.cn/showproblem.php?pid=2433

 

题意：

求删除任意一条边后，任意两点对的最短路之和

 

以每个点为根节点求一个最短路树，

只需要记录哪些边在最短路树上，记录整棵树的dis和

如果删除的边不在最短路树上，累加记录的dis和

否则，重新bfs求dis和

因为最短路树上有n-1条边，n棵树，所以只有（n-1）*n条边需要重新bfs

时间复杂度为n*n*m

 

求桥是 对面的low>自己的dfn，我求了一年假的桥

my god ！ ?(????)

 

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

using namespace std;

#define N 101
#define M 3001

int n,m;

int tot;
int front[N],nxt[M<<1],to[M<<1];

bool use[N][M],ok[M];
int sum[N],dis[N];
queue<int>q;

int dfn[N],low[N];
bool bridge[M];

void read(int &x)
{
    x=0; char c=getchar();
    while(!isdigit(c)) c=getchar();
    while(isdigit(c)) { x=x*10+c-‘0‘; c=getchar(); }
}

void add(int u,int v)
{
    to[++tot]=v; nxt[tot]=front[u]; front[u]=tot;
    to[++tot]=u; nxt[tot]=front[v]; front[v]=tot;
}

void tarjan(int u,int pre)
{
    dfn[u]=low[u]=++tot;
    for(int i=front[u];i;i=nxt[i])
    {
        if(i==(pre^1)) continue;
        if(!dfn[to[i]]) 
        {
            tarjan(to[i],i);
            low[u]=min(low[u],low[to[i]]);
            if(low[to[i]]>dfn[u]) bridge[i>>1]=true;
        }
        else low[u]=min(low[u],dfn[to[i]]);
    }
}

void bfs(int s)
{
    memset(dis,0,sizeof(dis));
    sum[s]=0;
    q.push(s);
    int now;
    while(!q.empty())
    {
        now=q.front();
        q.pop();
        for(int i=front[now];i;i=nxt[i])
            if(to[i]!=s && !dis[to[i]])
            {
                use[s][i>>1]=true;
                dis[to[i]]=dis[now]+1;
                sum[s]+=dis[to[i]];
                q.push(to[i]);
            }
    }
}

int bfs2(int s)
{
    int ans=0;
    memset(dis,0,sizeof(dis));
    q.push(s);
    int now;
    while(!q.empty())
    {
        now=q.front();
        q.pop();
        for(int i=front[now];i;i=nxt[i])
            if(ok[i>>1] && to[i]!=s && !dis[to[i]])
            {
                dis[to[i]]=dis[now]+1;
                ans+=dis[to[i]];
                q.push(to[i]);
            }
    }
    return ans;
}

void solve()
{
    
    int ans; 
    memset(ok,true,sizeof(ok));
    for(int i=1;i<=m;++i)
    {
        if(bridge[i]) 
        {
            puts("INF");
            continue;
        }
        ans=0; 
        for(int j=1;j<=n;++j)
            if(!use[j][i]) ans+=sum[j];
            else
            {
                ok[i]=false;
                ans+=bfs2(j);
                ok[i]=true;
            }
        printf("%d\n",ans);
    }
}

void clear()
{
    tot=1;
    memset(front,0,sizeof(front));
    memset(dfn,0,sizeof(dfn));
    memset(bridge,false,sizeof(bridge));
    memset(use,false,sizeof(use));
}

int main()
{
    int u,v;
    while(scanf("%d",&n)!=EOF)
    {
        clear();
        read(m);
        for(int i=1;i<=m;++i)
        {
            read(u); read(v);
            add(u,v);
        }
        for(int i=1;i<=n;++i) bfs(i);
        tot=0;
        tarjan(1,0);
        bool tag=false;
        for(int i=1;i<=n;++i)
            if(!dfn[i])
            {
                tag=true;
                break;
            }    
        if(!tag) solve();
        else 
        for(int i=1;i<=m;++i) puts("INF");
    }
}

Travel

Time Limit: 10000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3388    Accepted Submission(s): 1160

Problem Description

      One day, Tom traveled to a country named BGM. BGM is a small country, but there are N (N <= 100) towns in it. Each town products one kind of food, the food will be transported to all the towns. In addition, the trucks will always take the shortest way. There are M (M <= 3000) two-way roads connecting the towns, and the length of the road is 1.
      Let SUM be the total distance of the shortest paths between all pairs of the towns. Please write a program to calculate the new SUM after one of the M roads is destroyed.

 

 

Input

      The input contains several test cases.
      The first line contains two positive integers N, M. The following M lines each contains two integers u, v, meaning there is a two-way road between town u and v. The roads are numbered from 1 to M according to the order of the input.
      The input will be terminated by EOF.

 

 

Output

      Output M lines, the i-th line is the new SUM after the i-th road is destroyed. If the towns are not connected after the i-th road is destroyed, please output “INF” in the i-th line.

 

 

Sample Input

5 4 5 1 1 3 3 2 5 4 2 2 1 2 1 2

 

 

Sample Output

INF INF INF INF 2 2