/*
* Solution to IOI 2009 problem "raisins"
*
* This solution employs memoization in order to compute the the best
* way of cutting any sub-rectangle of the chocolate. To do this
* sufficiently quickly, we also precompute the number of raisins in
* each sub-rectangle subtended from the top-left corner of the chocolate.
*
* This allows us to compute the number of raisins in any rectangular
* sub-region of the chocolate -- consider the diagram below:
*
* (0,0)-------+-------(c,0)
* | A | B |
* +-------(a,b)-------+
* | C | D |
* (0,d)-------+-------(c,d)
*
* Suppose we wish to compute the number of raisins in the region
* (a,b) -> (c,d), D. Now D = (A + B + C + D) - (A + B) - (A + C) + A
* Each of these four quantities is given by the number of raisins
* in a rectangular sub-region whose top-left corner is (0,0).
*
* Carl Hultquist, [email protected]
*/
#include <iostream>
#include <cassert>
#include <climits>
#include <cstring>
using namespace std;
#define MAX_SIZE 50
#define MAX_VAL 1000
// The dimensions of the chocolate
int r, c;
// The number of raisins on each piece of chocolate
short chocolate[MAX_SIZE][MAX_SIZE];
// The number of raisins in a rectangular sub-region subtended from the
// top-left corner of the chocolate.
int raisins[MAX_SIZE][MAX_SIZE];
// The minimum payment that Bonny must make for Peter to cut any
// rectangular sub-region of the chocolate.
int best[MAX_SIZE][MAX_SIZE][MAX_SIZE][MAX_SIZE];
/**
* Computes the smallest cost of cutting up the rectangular portion of
* the chocolate from (r1,c1) to (r2,c2).
*/
int solve(int r1, int c1, int r2, int c2)
{
if (best[r1][c1][r2][c2] == -1)
{
// We haven't computed this previously, so compute it now.
if (r1 == r2 && c1 == c2) {
// If we have a single cell left, then we're done:
// there's no additional cost.
best[r1][c1][r2][c2] = 0;
}
else {
// We try all the possible cuts, and see which results in
// the smallest total payment.
int best_payment = INT_MAX;
// First try the rows
for (int r = r1 + 1; r <= r2; r++)
{
int payment = solve(r1, c1, r - 1, c2) + solve(r, c1, r2, c2);
if (payment < best_payment)
best_payment = payment;
}
// Now try the columns
for (int c = c1 + 1; c <= c2; c++)
{
int payment = solve(r1, c1, r2, c - 1) + solve(r1, c, r2, c2);
if (payment < best_payment)
best_payment = payment;
}
// Determine how many raisins
int base_raisins = raisins[r2][c2];
if (r1 > 0)
base_raisins -= raisins[r1 - 1][c2];
if (c1 > 0)
base_raisins -= raisins[r2][c1 - 1];
if (r1 > 0 && c1 > 0)
base_raisins += raisins[r1 - 1][c1 - 1];
best[r1][c1][r2][c2] = best_payment + base_raisins;
}
}
return best[r1][c1][r2][c2];
}
int main()
{
cin >> r >> c;
assert(1 <= r && r <= MAX_SIZE);
assert(1 <= c && c <= MAX_SIZE);
// Read in the number of raisins on each piece of chocolate
for (int i = 0; i < r; i++)
for (int j = 0; j < c; j++)
{
cin >> chocolate[i][j];
assert(0 <= chocolate[i][j] && chocolate[i][j] <= MAX_VAL);
}
memset(raisins, -1, sizeof(raisins));
memset(best, -1, sizeof(best));
// First precompute the number of raisins on each rectangular
// sub-region subtended from the top-left corner of the chocolate.
raisins[0][0] = chocolate[0][0];
for (int i = 1; i < r; i++)
raisins[i][0] = chocolate[i][0] + raisins[i - 1][0];
for (int j = 1; j < c; j++)
{
raisins[0][j] = chocolate[0][j] + raisins[0][j - 1];
for (int i = 1; i < r; i++)
raisins[i][j] = chocolate[i][j] + raisins[i - 1][j] + raisins[i][j - 1] - raisins[i - 1][j - 1];
}
// Find the minimum payment for cutting up the whole slab of
// chocolate.
cout << solve(0, 0, r - 1, c - 1) << endl;
return 0;
}
Compilation message
/tmp/ccjf11Hl.o: In function `main':
grader.cpp:(.text.startup+0x0): multiple definition of `main'
/tmp/ccYmeWIN.o:molecules.cpp:(.text.startup+0x0): first defined here
/tmp/ccjf11Hl.o: In function `main':
grader.cpp:(.text.startup+0x152): undefined reference to `find_subset(int, int, std::vector<int, std::allocator<int> >)'
collect2: error: ld returned 1 exit status
