| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 12566 | Accepted: 6061 |
Description
A university network is composed of N computers. System administrators gathered information on the traffic between nodes, and carefully divided the network into two subnetworks in order to minimize traffic between parts. A disgruntled computer science student Vasya, after being expelled from the university, decided to have his revenge. He hacked into the university network and decided to reassign computers to maximize the traffic between two subnetworks. Unfortunately, he found that calculating such worst subdivision is one of those PRoblems he, being a student, failed to solve. So he asks you, a more successful CS student, to help him. The traffic data are given in the form of matrix C, where Cij is the amount of data sent between ith and jth nodes (Cij = Cji, Cii = 0). The goal is to divide the network nodes into the two disjointed subsets A and B so as to maximize the sum ∑Cij (i∈A,j∈B).Input
The first line of input contains a number of nodes N (2 <= N <= 20). The following N lines, containing N space-separated integers each, represent the traffic matrix C (0 <= Cij <= 10000). Output file must contain a single integer -- the maximum traffic between the subnetworks.Output
Output must contain a single integer -- the maximum traffic between the subnetworks.Sample Input
30 50 3050 0 4030 40 0Sample Output
90Source
Northeastern Europe 2002, Far-Eastern Subregion題目意思：
有N臺電腦，給出一個N*N的鄰接矩陣，表示i節點到j節點的流量是Cij，且Cij=Cji。將N個網絡節點劃分為兩個不相交的子集A和B，計算最大的和ΣCij（i∈A，j∈B）。解題思路：
暴力枚舉+剪枝，借鑒了大神分組的方法。剪枝：只需要枚舉[2^(n-1)]-1次而不是n次。n個不同物品分成兩組的方法數是[2^(n-1)]-1，因為：ΣC(i,n)=2^n，其中i∈[0,n]；因為兩組均不為空，且C(0,n)=C(n,n)=1，所以ΣC(i,n)=(2^n)-2，其中i∈[1,n-1]；因為兩組相互區別，所以方法數是ΣC(i,n)/2,，即[2^(n-1)]-1，其中i∈[1,n-1]。#include<iostream>#include<cstdio>#include<iomanip>#include<cmath>#include<cstdlib>#include<cstring>#include<map>#include<algorithm>#include<vector>#include<queue>using namespace std;#define INF 0x3f3f3f3f#define MAXN 25int c[MAXN][MAXN];int vis[25];int main(){#ifdef ONLINE_JUDGE#else freopen("F:/cb/read.txt","r",stdin); //freopen("F:/cb/out.txt","w",stdout);#endif ios::sync_with_stdio(false); cin.tie(0); int n; while(cin>>n) { for(int i=0; i<n; i++) for(int j=0; j<n; j++) cin>>c[i][j];; memset(vis,0,sizeof(vis)); int ans=0; for(int i=0; i<(1<<(n-1)); i++)//枚舉2^n-1次就可以將所有情況枚舉出來 { ++vis[0]; for(int j=0; j<n; j++)//通過0,1的值將n個點分到兩個部分 { if(vis[j]==2) { vis[j]=0; ++vis[j+1]; } else break; } int res=0; for(int j=0; j<n; j++)//求出總和 if(vis[j])//計算從所有的1的所有的0的和 for(int k=0; k<n; k++) if(!vis[k]) res+=c[j][k]; ans=max(ans,res); } cout<<ans<<endl; } return 0;}/*30 50 3050 0 4030 40 0*/
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