dfs_poj_2531
2014-08-14 09:28
417 查看
题目链接:http://poj.org/problem?id=2531
Network Saboteur
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
Sample Output
基础的搜索题,用搜索穷举出各种状态,取最大值即可,搜索时基础,不难,计算最大值的时候,脑子抽了, 居然错了好几次
代码:
#include <stdio.h>
#include <string.h>
int map[22][22], n, Max;
short mark[22]; ///标记数组
void visit(int pos, int sum)
{
int i;
for( i = 1; i <= n; i++)
{
if(mark[i]){ sum -= map[pos][i]; } ///删除重复部分
else { sum += map[pos][i]; } ///添加
}
if(Max < sum)Max = sum;
for( i = pos + 1; i <= n; i++)
{
mark[i] = 1;
visit(i, sum);
mark[i] = 0;
}
}
int main()
{
int i, j;
while(~scanf("%d",&n))
{
Max = 0;
memset(map,0,sizeof(map));
memset(mark,0,sizeof(mark));
for( i = 1; i <= n; i++)
{
for( j = 1; j <= n; j++)
{
scanf("%d",&map[i][j]);
}
}
for( i = 1; i <= n; i++)
{
mark[i] = 1;
visit(i, 0); ///dfs 函数 写visit 写习惯了
mark[i] = 0;
}
printf("%d\n",Max);
}
return 0;
}
运行结果:
Network Saboteur
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 9246 | Accepted: 4345 |
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
3 0 50 30 50 0 40 30 40 0
Sample Output
90
基础的搜索题,用搜索穷举出各种状态,取最大值即可,搜索时基础,不难,计算最大值的时候,脑子抽了, 居然错了好几次
代码:
#include <stdio.h>
#include <string.h>
int map[22][22], n, Max;
short mark[22]; ///标记数组
void visit(int pos, int sum)
{
int i;
for( i = 1; i <= n; i++)
{
if(mark[i]){ sum -= map[pos][i]; } ///删除重复部分
else { sum += map[pos][i]; } ///添加
}
if(Max < sum)Max = sum;
for( i = pos + 1; i <= n; i++)
{
mark[i] = 1;
visit(i, sum);
mark[i] = 0;
}
}
int main()
{
int i, j;
while(~scanf("%d",&n))
{
Max = 0;
memset(map,0,sizeof(map));
memset(mark,0,sizeof(mark));
for( i = 1; i <= n; i++)
{
for( j = 1; j <= n; j++)
{
scanf("%d",&map[i][j]);
}
}
for( i = 1; i <= n; i++)
{
mark[i] = 1;
visit(i, 0); ///dfs 函数 写visit 写习惯了
mark[i] = 0;
}
printf("%d\n",Max);
}
return 0;
}
运行结果:
相关文章推荐
- POJ 2531 经典的DFS
- POJ 2531 Network Saboteur(DFS)
- POJ 2531 Network Saboteur(DFS & 随机化)解题报告
- poj 2531 DFS
- POJ 2531 Network Saboteur (dfs+水题)
- POJ 2531 dfs回溯(小剪枝)
- poj 2531 Network Saboteur (dfs)
- poj 2531 Network Saboteur (dfs)
- POJ 2531 - DFS - 经典
- poj-2531 Network Saboteur DFS
- POJ 2531 DFS + 递推
- POJ 2531-Network Saboteur(DFS)
- POJ-2531 Network Saboteur (DFS)
- POJ 2531 Network Saboteur(dfs)
- Network Saboteur poj 2531 dfs 简单搜索技巧和剪枝
- poj 2531(dfs)
- poj 2531 Network Saboteur( dfs )
- poj 2531 Network Saboteur(dfs)
- POJ 2531 Network revenger (DFS)
- POJ 2531(状态压缩||DFS)