NEU 1685: All Pair Shortest Path

题目描述

Bobo has a directed graph G with n vertex labeled by 1,2,3,..n.

Let D(i,j) be the number of edges from vertex i to vertex j on the shortest path.

If the shortest path does not exist,then D(i,j)=n.

Bobo would like to find the sum of D(i,j)*D(i,j) for all 1<=i<=n and 1<=j<=n.

输入

There are no more than 5 test cases.

The first line contains an integer n(1<=n<=1000).

The i-th of the following n lines contains n integers g(i,1),g(i,2),..g(i,n).

If there is an edge from i to j,then g(i,j)=1,otherwise g(i,j)=0;

输出

An integer denotes the sum of D(i,j)*D(i,j) for all 1<=i<=n and 1<=j<=n.

样例输入

样例输出

15

8
题意就是求所有D（i，j）*D（i，j）的和。D(i,j)代表i j之间的最短路径。
正常的想法肯定是 bfs求出任意两点之间的最短路径  但这样做的时间复杂度大概n^3  会超时。
得优化。用set维护未访问的点  因为set的删除 插入的操作都是logn 而n最大是1000。所以总体的时间复杂度是  常数*n^2

/* ***********************************************

Author        :guanjun

Created Time  :2016/3/21 16:44:25

File Name     :neu1685.cpp

************************************************ */

#include <iostream>

#include <cstring>

#include <cstdlib>

#include <stdio.h>

#include <algorithm>

#include <vector>

#include <queue>

#include <set>

#include <map>

#include <string>

#include <math.h>

#include <stdlib.h>

#include <iomanip>

#include <list>

#include <deque>

#include <stack>

#define ull unsigned long long

#define ll long long

#define mod 90001

#define INF 0x3f3f3f3f

#define maxn 1010

#define cle(a) memset(a,0,sizeof(a))

const ull inf = 1LL << ;

const double eps=1e-;

using namespace std;

priority_queue<int,vector<int>,greater<int> >pq;

struct Node{

    int x,y;

};

struct cmp{

    bool operator()(Node a,Node b){

        if(a.x==b.x) return a.y> b.y;

        return a.x>b.x;

    }

};

bool cmp(int a,int b){

    return a>b;

}

int n;

char mp[maxn][maxn];

int vis[maxn];

int dis[maxn];

ll sum=;

set<int>s;

set<int>::iterator it;

void solve(){

    cle(dis);

    s.clear();

    int cnt=;

    queue<int>q;

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

        q.push(i);

        for(int j=;j<=n;j++){

            if(i==j)continue;

            if(mp[i][j]=='')dis[j]=,q.push(j);

            else s.insert(j);

        }

        //cout<<"s "<<s.size()<<endl;

        while(!q.empty()){

            int x=q.front();q.pop();

            cnt=;

            for(it=s.begin();it!=s.end();it++){

                if(mp[x][*it]==''){

                    q.push(*it);

                    dis[*it]=dis[x]+;

                    vis[++cnt]=*it;

                }

            }

            for(int j=;j<=cnt;j++)s.erase(vis[j]);

        }

        for(int j=;j<=n;j++){

            if(i==j)continue;

            else{

                if(dis[j]>)sum+=dis[j]*dis[j];

                else sum+=n*n;

            }

        }

        cle(dis);

    }

}

int main()

{

    #ifndef ONLINE_JUDGE

    freopen("in.txt","r",stdin);

    #endif

    //freopen("out.txt","w",stdout);

    while(cin>>n){

        sum=;

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

            scanf("%s",mp[i]+);

        }

        /*

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

            for(int j=1;j<=n;j++)

                printf("%c%c",mp[i][j],j==n?10:' ');

        }*/

        solve();

        printf("%lld\n",sum);

    }

    return ;

}