POJ 1682 Clans on the Three Gorges

Description
On the banks of the Three Gorges of the Yangtze River live some old clans. For historical reasons, clans on the same gorge hate each other, and conflicts often arise between them for lands and recourses. On the contrary they have a long and deep friendship with the clans on the other gorges. But it’s difficult for them to travel to the other sides of the gorge because of the torrent flow of the Yangtze River.
One day they decide to build suspension bridges between the gorges. The gorges have three banks (figure 1), and each bank is rolling so clans have different altitude. Every clan will not share a same bridge with other clans on the same bank, so there must be at least one bridge for each clan. And for safety reason, bridges are not allowed to cross each other overhead, or if a bridge collapsed, it would fall on another one. Experts point out that if the altitudes of the two ends of a bridge are different, the bridge will be troublesome to build and cost more. It’s estimated that the cost of a bridge is equal to the fall of the ends of the bridge. The clans are not rich, so they employ you to find a solution to minimize the cost to build the suspension bridges to satisfy their demand.
As show in figure 1, the three gorges are named X, Y and Z respectively. A clan is represented by a cycle, with its name and altitude in. Clans on gorge X are named x1, x2… xn in counter-clockwise order. Those on gorge Y and gorge Z are y1, y2… ym and z1, z2? zk in counter-clockwise order too. A bridge is represented by a line between two clans with its cost on it. No two lines can intersect and no clan has zero degree according to the description of this problem.
For example, figure 2 is invalid solution because the two bridges intersect. Figure 3 is also invalid for the reason that clan y1 doesn’t has a bridge for its own. Figure 1 and Figure 4 both are valid solutions for there are no intersected bridges or isolated clan in these solutions. The cost of a solution is the sum of all the costs of the bridges in the solution. And the cost of a bridge is equal to the absolute value of the difference of the heights of the two ends. So figure 1 has cost 11 against 14 of figure 4. It’s clear that figure 1 is a better solution. In fact, it’s one of the best solutions. You are required to write a program to find the minimum cost.

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【题目分析】
蒟蒻看到之后不会做，只会解决二维的问题。然后就借鉴了别人的题解（说的好听，就是ctrl+c ctrl+v）。这道题的思路大概就是这样：
1、两两解决，共三组。dp[i][j]表示一面匹配到i，另一面匹配到j的情况。
2、将这三组合起来。
合并的时候由于不能交叉所以只有几种情况，然后写出来（找到最小值），就可以了。

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【代码】

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <cmath>
using namespace std;
int n,m,p,a[150],b[150],c[150],xz[150][150],xy[150][150],yz[150][150],ans,tt;
void read()
{
    scanf("%d%d%d",&m,&n,&p);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    for(int i=1;i<=m;i++) scanf("%d",&b[i]);
    for(int i=1;i<=p;i++) scanf("%d",&c[i]);
    for(int i=0;i<150;i++)
        for(int j=0;j<150;j++)
            xz[i][j]=xy[i][j]=yz[i][j]=999999;
    ans=999999;
}

void go()
{
    xz[0][p+1]=0;
    for(int i=1;i<=n;i++)
        for(int j=p;j>=1;j--)
            xz[i][j]=min(min(xz[i-1][j+1],xz[i-1][j]),xz[i][j+1])+abs(a[i]-c[j]);
    xy[n+1][0]=0;
    for(int i=n;i>=1;i--)
        for(int j=1;j<=m;j++)
            xy[i][j]=min(min(xy[i+1][j-1],xy[i+1][j]),xy[i][j-1])+abs(a[i]-b[j]);
    yz[m+1][0]=0;
    for(int i=m;i>=1;i--)
        for(int j=1;j<=p;j++)
            yz[i][j]=min(min(yz[i+1][j-1],yz[i+1][j]),yz[i][j-1])+abs(b[i]-c[j]);
    for(int i=0;i<=n+1;i++)
        for(int j=0;j<=m+1;j++)
            for(int k=0;k<=p+1;k++)
            {
                ans=min(ans,xy[i][j]+xz[i][k]+yz[j][k]);
                ans=min(ans,xy[i][j]+xz[i][k]+yz[j+1][k]);  
                ans=min(ans,xy[i+1][j]+xz[i][k]+yz[j][k]);  
                ans=min(ans,xy[i][j]+xz[i][k+1]+yz[j][k]);
                ans=min(ans,xy[i+1][j]+xz[i][k]+yz[j+1][k]);  
                ans=min(ans,xy[i][j]+xz[i][k+1]+yz[j+1][k]);  
                ans=min(ans,xy[i+1][j]+xz[i][k+1]+yz[j][k]);
                ans=min(ans,xy[i+1][j]+xz[i][k+1]+yz[j+1][k]);
            }
    printf("%d\n",ans);
}

int main()
{
    scanf("%d",&tt);
    while(tt--)
    {
        read();
        go();
    }
    return 0;
}