uva 10549 Russian Dolls (DP)

Problem D : Russian Dolls
From:UVA, 10549

Problem A: Russian Dolls

Russian nesting dolls are brightly painted hollowwooden figures. The dolls in a set have roughly the same shape, typically humanoid, but different sizes.When the set is assembled, the biggest doll containsthe second-biggest doll, the second-biggest containsthe
third-biggest, and so on.
We can approximate the shape of a doll as a cylinder of height h, diameter
d, and wall thickness w. Such a doll would have ahollow of height
h-2w and diameter d-2w.

Boris and Natasha each has a set of dolls. Thesets are nearly identical; each has the same numberof dolls, which look the same but differ in their dimensions.Last night Boris and Natasha were playing with their dolls andleft them in the living room. Their
mother tidiedthem away, dumping them all in one box. Can youhelp Boris and Natasha separate their sets of dolls?

Standard Input will consist of several test cases. Thefirst line of each test case will contain
n, thenumber of dolls in each set (1 < n <= 100). 2nlines follow; each gives the dimensions,
h, d,w of a different doll (h,d >= 2w > 0). A line containing 0 follows the last test case.

For each test case, separate the dolls into two sets ofnesting dolls such that, within each set, the dolls fitwithin each other, standing straight up, as described above. The first
n lines of output should give thedimensions of the dolls in one set, in decreasing order byheight. The next line should contain a single hyphen, "-".The next
n lines should give the dimensionsof the dolls in the second set, also in decreasing order by height. There will always be asolution. If there are many solutions,any will do. Output an empty line betweentest cases.

Sample Input

3
100 100 3
97 97 3
94 94 3
91 91 3
88 88 3
85 85 3
5
100 100 1
97 97 3
98 98 1
96 96 1
94 94 1
92 92 1
90 90 1
88 88 1
86 86 1
84 84 1
0

Possible Output

100 100 3
94 94 3
88 88 3
-
97 97 3
91 91 3
85 85 3

100 100 1
98 98 1
96 96 1
94 94 1
92 92 1
-
97 97 3
90 90 1
88 88 1
86 86 1
84 84 1

[code]#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
#include <stack>
using namespace std;

const int maxn = 101;
bool dp[2*maxn][2*maxn][maxn][maxn];
short vis[2*maxn][2*maxn][maxn][maxn] = {0};
stack<int> v1 , v2;
struct DOLL{
int h , d , w;
DOLL(int a = 0 , int b = 0,int c = 0){
h = a , d = b , w = c;
}
}Doll[2*maxn];
int n;
short v;

bool cmp(DOLL d1 , DOLL d2){
if(d1.h-2*d1.w == d2.h-2*d2.w){
return d1.d-2*d1.w < d2.d-2*d2.w;
}
return d1.h-2*d1.w < d2.h-2*d2.w;
}

void readcase(){
int h , d , w;
for(int i = 1;i <= 2*n;i++){
scanf("%d%d%d" , &h , &d , &w);
Doll[i] = DOLL(h , d , w);
}
}

bool check(int a , int b){
if(Doll[a].d-2*Doll[a].w >= Doll[b].d && Doll[a].h-2*Doll[a].w >= Doll[b].h){
return true;
}
//cout << a << "::::" << b<<"=" << Doll[a].h << " " << Doll[b].h << endl;
return false;
}

bool DP(int k , int l , int r){
//cout << k <<":" << l << " " << r << endl;
if(k >= 2*n+1){
//cout << v1.size() << v2.size() << endl;
if(v1.size() == v2.size()){
return true;
}
return false;
}
//cout << k <<":" << l << " " << r << endl;
if(vis[l][r][v1.size()][v2.size()] == v) return dp[l][r][v1.size()][v2.size()];
if(l == 0 || check(k , v1.top())){
v1.push(k);
if(DP(k+1 , k , r )){
return dp[l][r][v1.size()-1][v2.size()] = true;
}
v1.pop();
}
if(r == 0 || check(k , v2.top())){
v2.push(k);
if(DP(k+1 , l , k )){
return dp[l][r][v1.size()][v2.size()-1] = true;
}
v2.pop();
}
vis[l][r][v1.size()][v2.size()] = v;
return dp[l][r][v1.size()][v2.size()] = false;
}

void computing(){
sort(Doll+1 , Doll+2*n+1 , cmp);
DP(1 , 0 , 0);
while(!v2.empty()){
int t = v2.top();
v2.pop();
printf("%d %d %d\n" , Doll[t].h , Doll[t].d , Doll[t].w);
}
printf("-\n");
while(!v1.empty()){
int t = v1.top();
v1.pop();
printf("%d %d %d\n" , Doll[t].h , Doll[t].d , Doll[t].w);
}
}

int main(){
//freopen("in" , "r" , stdin);
v = 1;
while(scanf("%d" , &n) == 1 && n){
if(v != 1){
printf("\n");
}
readcase();
computing();
v++;
}
return 0;
}

[/code]