在这里插入图片描述
UVA1629
定义
$UpperLeft$为矩形左上角,$LowerRight$为右下角。
$dp[UpperLeft.x][UpperLeft.y][LowerRight.x][LowerRight.y]$为由两组坐标确定的矩形区域最少要切割多少长度才能满足要求。
$PrefixSum[i][j]$为从(0,0)到(i,j)的矩形区域一共有多少樱桃。(用于O(1)计算出给定矩形区域内樱桃数量)。
初始化

转移方程

AC代码

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#include<iostream>
#include<string>
#include<cstring>
#include<algorithm>
#include<vector>
#include<cmath>
#include<map>
using namespace std;
int n, m, k;
bool HasCherry[21][21];
int PrefixSum[22][22];
int dp[22][22][22][22];
constexpr static int inf = 0x3f3f3f3f;
struct Point {
int x, y;
};
bool Input() {
if (scanf("%d%d%d", &n, &m, &k) == EOF) {
return false;
}
memset(HasCherry, false, sizeof(HasCherry));
while (k--) {
int x, y;
scanf("%d%d", &x, &y);
HasCherry[x][y] = true;
}
return true;
}
void InitPrefixSum() {
memset(PrefixSum, 0x0, sizeof(PrefixSum));
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
PrefixSum[i][j] = PrefixSum[i - 1][j] + PrefixSum[i][j - 1] - PrefixSum[i - 1][j - 1] + HasCherry[i][j];
}
}
}
const int getCherryNumber(const Point& UpperLeft,const Point& LowerRight) {
return
PrefixSum[LowerRight.x][LowerRight.y]
- PrefixSum[LowerRight.x][UpperLeft.y - 1]
- PrefixSum[UpperLeft.x - 1][LowerRight.y]
+ PrefixSum[UpperLeft.x - 1][UpperLeft.y - 1];
}
int DP(Point UpperLeft, Point LowerRight) {
int& Ans = dp[UpperLeft.x][UpperLeft.y][LowerRight.x][LowerRight.y];
const int&& CherryNumber = getCherryNumber(UpperLeft, LowerRight);
//樱桃数量为0,不符合题意,返回inf
if (!CherryNumber) {
return Ans = inf;
}
//不需要切了,返回0
else if (CherryNumber == 1) {
return Ans = 0;
}
if (Ans != inf) {
return Ans;
}
//横着切,枚举每一种切割方案
for (int i = UpperLeft.x; i < LowerRight.x; ++i) {
Ans = min(
Ans,
DP(UpperLeft, { i,LowerRight.y })
+ DP({ i + 1,UpperLeft.y }, LowerRight)
+ LowerRight.y - UpperLeft.y + 1
);
}
//竖着切
for (int i = UpperLeft.y; i < LowerRight.y; ++i) {
Ans = min(
Ans,
DP(UpperLeft, { LowerRight.x,i })
+ DP({ UpperLeft.x,i + 1 }, LowerRight)
+ LowerRight.x - UpperLeft.x + 1
);
}
return Ans;
}
int main() {
int&& Case = 0;
while (Input()) {
InitPrefixSum();
memset(dp, 0x3f, sizeof(dp));
printf("Case %d: %d\n", ++Case, DP({ 1,1 }, { n, m }));
}
return 0;
}