Codeforces Beta Round #25 (Div. 2 Only) C. Roads in Berland

Floyd算法变种

最新推荐文章于 2022-12-16 16:19:21 发布

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最新推荐文章于 2022-12-16 16:19:21 发布

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原文链接：http://www.cnblogs.com/pk28/p/4755736.html

本文介绍了一种基于Floyd算法的变种应用，用于解决在给定图中进行多次边权更新后，任意两点间最小距离之和的问题。通过巧妙地优化路径查找过程，避免了重复计算和超时问题。

There are n cities numbered from 1 to n in Berland. Some of them are connected by two-way roads. Each road has its own length — an integer number from 1 to 1000. It is known that from each city it is possible to get to any other city by existing roads. Also for each pair of cities it is known the shortest distance between them. Berland Government plans to build k new roads. For each of the planned road it is known its length, and what cities it will connect. To control the correctness of the construction of new roads, after the opening of another road Berland government wants to check the sum of the shortest distances between all pairs of cities. Help them — for a given matrix of shortest distances on the old roads and plans of all new roads, find out how the sum of the shortest distances between all pairs of cities changes after construction of each road.

Input

The first line contains integer n (2 ≤ n ≤ 300) — amount of cities in Berland. Then there follow n lines with n integer numbers each — the matrix of shortest distances. j-th integer in the i-th row — di, j, the shortest distance between cities i and j. It is guaranteed thatdi, i = 0, di, j = dj, i, and a given matrix is a matrix of shortest distances for some set of two-way roads with integer lengths from 1 to 1000, such that from each city it is possible to get to any other city using these roads.

Next line contains integer k (1 ≤ k ≤ 300) — amount of planned roads. Following k lines contain the description of the planned roads. Each road is described by three space-separated integers ai, bi, ci (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ ci ≤ 1000) — ai and bi — pair of cities, which the road connects, ci — the length of the road. It can be several roads between a pair of cities, but no road connects the city with itself.

Output

Output k space-separated integers qi (1 ≤ i ≤ k). qi should be equal to the sum of shortest distances between all pairs of cities after the construction of roads with indexes from 1 to i. Roads are numbered from 1 in the input order. Each pair of cities should be taken into account in the sum exactly once, i. e. we count unordered pairs.

Sample test(s)

input

output

input

output

17 12

题目大意：给出一个矩阵，表示一个图和边权值（这里的边权值已经是两点间的最短距离）。然后给出K次修改，对于每一次边权修改，输出修改后任意两点的最小距离的和。

Floyd变形。首先任意两点的最小距离的和只能暴力求解，对于边的更新，我们肯定会想到跑一边floyd但这会超时。

对于第修改的边a,b我们可以知道任意两点i，j的最短距离要么经过a-b，要么经过b-a，要么不经过a点也不经过b点。比n^3要好一些.

dis[i][j]=dis[j][i]=min(dis[i][j],dis[i][a]+dis[b][j]+c);
dis[i][j]=dis[j][i]=min(dis[i][j],dis[i][b]+dis[a][j]+c);

/* ***********************************************
Author        :pk29
Created Time  :2015/8/23 8:53:43
File Name     :4.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 10000+10
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;

bool cmp(int a,int b){
    return a>b;
}
int dis[310][310];
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int n,k;
    cin>>n;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++){
            scanf("%d",&dis[i][j]);
        }
    cin>>k;
    int a,b,c;
    while(k--){
        scanf("%d%d%d",&a,&b,&c);
        if(c<dis[a][b]){
            for(int i=1;i<=n;i++)
                for(int j=1;j<=n;j++){
                    dis[i][j]=dis[j][i]=min(dis[i][j],dis[i][a]+dis[b][j]+c);
                    dis[i][j]=dis[j][i]=min(dis[i][j],dis[i][b]+dis[a][j]+c);
                }
        }
        ll sum=0;
        for(int i=1;i<=n;i++)
            for(int j=i+1;j<=n;j++){
                sum+=dis[i][j];
            }
        cout<<sum<<" ";
    }
    return 0;
}