http://codeforces.com/problemset/problem/295/B

题意:

给定一个有边权的有向图。再给定一个1~n的排列。
按排列中的顺序依次删除点,问每次删除后,所有点对的最短路的和是多少。
 

删点看做倒序加点,然后模拟一遍Floyd

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm> using namespace std; typedef long long LL; #define N 501 int a[N]; LL f[N][N];
LL ans[N]; bool use[N]; template<typename T>
void read(T &x)
{
x=; char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) { x=x*+c-''; c=getchar(); }
} int main()
{
int n;
read(n);
for(int i=;i<=n;++i)
for(int j=;j<=n;++j)
read(f[i][j]);
for(int i=;i<=n;++i) read(a[i]);
int k;
for(int g=n;g;--g)
{
use[a[g]]=true;
k=a[g];
for(int i=;i<=n;++i)
for(int j=;j<=n;++j)
if(f[i][k]!=- && f[j][k]!=-)
f[i][j]=f[i][j]==- ? f[i][k]+f[k][j] : min(f[i][j],f[i][k]+f[k][j]);
for(int i=;i<=n;++i)
for(int j=;j<=n;++j)
if(use[i] && use[j]) ans[g]+=f[i][j];
}
for(int i=;i<=n;++i) printf("%I64d ",ans[i]);
}
B. Greg and Graph
time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Greg has a weighed directed graph, consisting of n vertices. In this graph any pair of distinct vertices has an edge between them in both directions. Greg loves playing with the graph and now he has invented a new game:

  • The game consists of n steps.
  • On the i-th step Greg removes vertex number xi from the graph. As Greg removes a vertex, he also removes all the edges that go in and out of this vertex.
  • Before executing each step, Greg wants to know the sum of lengths of the shortest paths between all pairs of the remaining vertices. The shortest path can go through any remaining vertex. In other words, if we assume that d(i, v, u) is the shortest path between vertices v and u in the graph that formed before deleting vertex xi, then Greg wants to know the value of the following sum: .

Help Greg, print the value of the required sum before each step.

Input

The first line contains integer n (1 ≤ n ≤ 500) — the number of vertices in the graph.

Next n lines contain n integers each — the graph adjacency matrix: the j-th number in the i-th line aij (1 ≤ aij ≤ 105, aii = 0) represents the weight of the edge that goes from vertex i to vertex j.

The next line contains n distinct integers: x1, x2, ..., xn (1 ≤ xi ≤ n) — the vertices that Greg deletes.

Output

Print n integers — the i-th number equals the required sum before the i-th step.

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64dspecifier.

Examples
input
1
0
1
output
0 
input
2
0 5
4 0
1 2
output
9 0 
input
4
0 3 1 1
6 0 400 1
2 4 0 1
1 1 1 0
4 1 2 3
output
17 23 404 0 

Codeforces 295 B. Greg and Graph的更多相关文章

  1. ural 1091. Tmutarakan Exams 和 codeforces 295 B. Greg and Graph

    ural 1091 题目链接:http://acm.timus.ru/problem.aspx?space=1&num=1091 题意是从1到n的集合里选出k个数,使得这些数满足gcd大于1 ...

  2. ACM - 最短路 - CodeForces 295B Greg and Graph

    CodeForces 295B Greg and Graph 题解 \(Floyd\) 算法是一种基于动态规划的算法,以此题为例介绍最短路算法中的 \(Floyd\) 算法. 我们考虑给定一个图,要找 ...

  3. 那些年我们写过的三重循环----CodeForces 295B Greg and Graph 重温Floyd算法

    Greg and Graph time limit per test 3 seconds memory limit per test 256 megabytes input standard inpu ...

  4. [CodeForces - 296D]Greg and Graph(floyd)

    Description 题意:给定一个有向图,一共有N个点,给邻接矩阵.依次去掉N个节点,每一次去掉一个节点的同时,将其直接与当前节点相连的边和当前节点连出的边都需要去除,输出N个数,表示去掉当前节点 ...

  5. CodeForces 295B Greg and Graph (floyd+离线)

    <题目链接> 题目大意:给定$n$个点的有向完全带权图$(n\leq500)$,现在进行$n$次操作,每次操作从图中删除一个点(每删除一个点,都会将与它相关联的边都删除),问你每次删点之前 ...

  6. Codeforces 715B & 716D Complete The Graph 【最短路】 (Codeforces Round #372 (Div. 2))

    B. Complete The Graph time limit per test 4 seconds memory limit per test 256 megabytes input standa ...

  7. Codeforce 295B Greg and Graph(Floyd的深入理解)

    题目链接:http://codeforces.com/problemset/problem/295/B 题目大意:给出n个点的完全有权有向图,每次删去一个点,求删掉该点之前整张图各个点的最短路之和(包 ...

  8. codeforces 715B:Complete The Graph

    Description ZS the Coder has drawn an undirected graph of n vertices numbered from 0 to n - 1 and m ...

  9. Codeforces 1082 G - Petya and Graph

    G - Petya and Graph 思路: 最大权闭合子图 对于每条边,如果它选了,那么它连的的两个点也要选 边权为正,点权为负,那么就是求最大权闭合子图 代码: #pragma GCC opti ...

随机推荐

  1. 7、Docker监控方案(cAdvisor+InfluxDB+Grafana)

    一.组件介绍 我们采用现在比较流行的cAdvisor+InfluxDB+Grafana组合进行Docker监控. 1.cAdvisor(数据采集) 开源软件cAdvisor(Container Adv ...

  2. Flutter - 给App增加启动屏幕(Splash Screen)并且设置背景颜色

    先看一下效果图,启动图最好设置为png格式的透明图,以防图片填充不满的时候背景图会非常的煞白(Flutter 默认背景色是白色). 打开android\app\src\main\res\drawabl ...

  3. 使用ClosedXML,读取到空行

    最近项目中使用了ClosedXML.dll来处理Excel,在读取Excel的时候,用workSheet.Rows()获取Excel行数,默认读取Excel最大行数1048576 所以为了读取到不是空 ...

  4. python基础篇----基本数据类型

    bit  #bit_length 当前数字的二进制,只用用n位来表示a = 123b = a.bit_length()print(b)#==>7

  5. Unity Inspector添加自定义按钮(Button)

    在Unity开发游戏的时候,为了有一个更快更方便的工作流,我们往往会在Editor下开发一些方便实用的工具.在工具中,用到最多,最关键的就是按钮,它是工具的首席执行官.下面就用最简单的代码来演示添加一 ...

  6. Unity攻击敌人时产生泛白效果

    Shader的代码如下,主要是将透明度为1的像素点输出为白色,其中_BeAttack表示角色被攻击的泛白状态 // Upgrade NOTE: replaced 'mul(UNITY_MATRIX_M ...

  7. 部署jar项目常用命令

      netstat -tunlp | grep  ××   查询出端口为××在运行应用的线程ip   kill -9  ××     关闭线程ip 为 ××的应用   rm  -f  ××.jar  ...

  8. FINAUNCE金融业增速反弹信贷投放创新高叠加股市回暖

    FINAUNCE金融业增速反弹信贷投放创新高叠加股市回暖,金融业增加值增速回暖,不过难以回到2015年的巅峰. 国家统计局4月18日发布的数据显示,今年一季度,国内生产总值21.34万亿元,按可比价格 ...

  9. Cloud Native Weekly | KubeCon首登中国,华为云亮相KubeCon 2018,微软云服务又罢工

    1.KubeCon首登中国,Kubernetes将如何再演进? 11月14日,由CNCF发起的云原生领域全球最大的峰会之一KubeCon+CloudNativeCon首次登陆中国,中国已经成为云原生领 ...

  10. 词频统计 SPEC 20160911

    本文档随时可能修改,并且没有另行通知. 请确保每一次在开始修改你的代码前,读标题中的日期,如果晚于你上次阅读, 请重读一次. 老五在寝室吹牛他熟读过<鲁滨逊漂流记>,在女生面前吹牛热爱&l ...