传送门
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 %I64d specifier.
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
- 总结;这道题其实就是用的Floyd，但是如果直接使用Floyd复杂度就是n^4，肯定不行，但是我们换一个角度想一下弗洛伊德的实现原理，就是在两个点之间选择一个中间的点然后过去，而在这个题里面，我们是一次次去掉一个顶点，然后我们就可以反过来思考，每一次加入一个点，这里我们加入这个点的时候，我们就可以把它当做那个中间点来更新其他的路径，然后我们在算每一次的总路径长度的时候我们需要判断一下这个顶点是否在图里面（前面已经说了我们从后往前每次加入一个点），把答案记录在一个数组里面，然后把这个数组输出就行了
这个题要用 long long
由于很菜，避免小错误直接把里面所有的整形变量全部改成long long了
#include <iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn = 510;
const int inf = 0x3f3f3f3f;
ll mp[maxn][maxn];
int main( ){
ll n,cnt=0;
ll ask[maxn];
ll vis[maxn];
ll ans[maxn];
scanf("%lld",&n);
memset(vis,0,sizeof(vis));
for(ll i=1;i<=n;i++){
for(ll j=1;j<=n;j++){
scanf("%lld",&mp[i][j]);
}
}
for(ll i=1;i<=n;i++){
scanf("%lld",&ask[i]);
}
for(ll k=n;k>=1;k--){
ll now=ask[k];//把这个点作为中间点
vis[now]=1;
for(ll i=1;i<=n;i++){
for(ll j=1;j<=n;j++){
mp[i][j]=min(mp[i][now]+mp[now][j],mp[i][j]);//用它去更新其他的路径
}
}
ll sum=0;
for(ll i=1;i<=n;i++){
for(ll j=1;j<=n;j++){
if(vis[i]&&vis[j]){
sum+=mp[i][j];
}
}
}
ans[cnt++]=sum;
}
for(ll i=n-1;i>=0;i--){
printf("%lld%c",ans[i],i==0?'\n':' ');
}
}