POJ 1636 Prison rearrangement DFS+0/1背包
题目链接： POJ 1636 Prison rearrangement
Prison rearrangement
| Time Limit: 3000MS | Memory Limit: 10000K | |
| Total Submissions: 2194 | Accepted: 984 |
Description
In order to lower the risk of riots and escape attempts, the boards of two nearby prisons of equal prisoner capacity, have decided to rearrange their prisoners among themselves. They want to exchange half of the prisoners of one prison, for half of the prisoners
of the other. However, from the archived information of the prisoners' crime history, they know that some pairs of prisoners are dangerous to keep in the same prison, and that is why they are separated today, i.e. for every such pair of prisoners, one prisoners
serves time in the first prison, and the other in the second one. The boards agree on the importance of keeping these pairs split between the prisons, which makes their rearrangement task a bit tricky. In fact, they soon find out that sometimes it is impossible
to fulfil their wish of swapping half of the prisoners. Whenever this is the case, they have to settle for exchanging as close to one half of the prisoners as possible.
Input
On the first line of the input is a single positive integer n, telling the number of test scenarios to follow. Each scenario begins with a line containing two non-negative integers m and r, 1 < m < 200 being the number of prisoners in each of the two prisons,
and r the number of dangerous pairs among the prisoners. Then follow r lines each containing a pair xi yi of integers in the range 1 to m,which means that prisoner xi of the first prison must not be placed in the same prison as prisoner yi of the second prison.
Output
For each test scenario, output one line containing the largest integer k <= m/2 , such that it is possible to exchange k prisoners of the first prison for k prisoners of the second prison without getting two prisoners of any dangerous pair in the same prison.
Sample Input
3 101 0 3 3 1 2 1 3 1 1 8 12 1 1 1 2 1 3 1 4 2 5 3 5 4 5 5 5 6 6 7 6 8 7 8 8
Sample Output
50 0 3
Source
题意：
有两个监狱都有m个囚犯，如今为了更好地管理，须要相互交换一部分人（小于等于m/2）。可是有一个问题就是，某一些有冲突的人的人不能待在一个监狱里面。也就是说。监狱1中的囚犯A和监狱2的囚犯B假设放在一起的话，将会产生非常严重的后果。如今你要求出可行的最大交换人数（不大于一半）。使得有冲突的人不能待在一个监狱里面。
思路：
这道题卡了挺久，网上看了非常多题解，然后自己认真想了想，发觉自己对背包的理解还不是非常透彻。
想想看，首先要将有关系的人分开成一个个集合（p, q)。表示监狱1中要拿出p个人交换的话，监狱2得同一时候拿出q个人。怎样找到这个集合呢？细心地人一眼就看出了是求连通分量了。
先构图，由于两个监狱的囚犯的人数和编号都是一样的，为了构图方便。我们将第二个监狱的囚犯的便后向后挪m（+m)。
然后有关系的两个人之间连一条无向边。之后的事情就交给dfs求连通分支了，当然中间要记录监狱1和监狱2须要交换多少个人。
那如今我们得到了cnt个连通分支了，也就是有cnt个集合（p, q)，如今看看0/1背包的思想。
我们用二维数组来记录在不发生危急的情况下可进行交换的情况。dp[i][j] = true表示第一个监狱拿出i个人、第二个监狱拿出j个人进行交换不产生冲突的情况。
利用滚动数组的思想，从后往前更新全部可能情况。
由于最后要保证两个监狱交换人数同样。因而找到最大的i使dp[i][i] == true，i（《= m/2）就是我们要求的结果。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int m, g[402][402];
int p1[202], p2[202], dp[110][110];
int cnt;
bool vis[402];
void dfs(int s)
{
vis[s] = true;
if(s <= m)
p1[cnt]++;
else
p2[cnt]++;
for(int i = 1; i <= 2*m; i++)
if(!vis[i] && g[s][i])
dfs(i);
}
int main()
{
int t, r, a, b;
//freopen("out.txt", "w", stdout);
scanf("%d", &t);
while(t--)
{
scanf("%d%d", &m, &r);
memset(g, 0, sizeof(g));
while(r--)
{
scanf("%d%d", &a, &b);
g[a][b+m] = g[b+m][a] = 1;
}
memset(vis, false, sizeof(vis));
cnt = 0;
for(int i = 1; i <= 2*m; i++)
if(!vis[i])
{
p1[cnt]= p2[cnt] = 0;
dfs(i);
cnt++;
}
memset(dp, false, sizeof(dp));
dp[0][0] = true;
for(int i = 0; i < cnt; i++)
for(int j = m/2; j >= p1[i]; j--)
for(int k = m/2; k >= p2[i]; k--)
if(dp[j-p1[i]][k-p2[i]])
dp[j][k] = true;
int index = m/2;
for(int i = m/2; i >= 0; i--)
if(dp[i][i])
{
index = i;
break;
}
printf("%d\n", index);
}
return 0;
}
