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
3
010
001
100
2
10
01
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 ;
}
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