1629 - Cake slicing

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https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=848&problem=4504&mosmsg=Submission+received+with+ID+16048725
我有话说：
这道题的大致思想是动态规划中的记忆化搜索。求对于给定区域（r1,c1,r2,c2)达到题目目标的最少的切割线总长度。一步一步划分下去。最后得到答案。注意数组dp不宜过大。

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <vector>#include <queue>using namespace std;const int maxn=100;const int INF=0x3f3f3f3f;int n,m,k;int G[maxn][maxn];int dp[maxn][maxn][maxn][maxn];void Init(){    for(int i=2;i<=n;i++){        for(int j=1;j<=m;j++){            G[i][j]+=G[i-1][j];        }    }    for(int j=2;j<=m;j++){        for(int i=1;i<=n;i++){            G[i][j]+=G[i][j-1];        }    }}int count_num(int r1,int c1,int r2,int c2){    return G[r2][c2]-G[r2][c1-1]-G[r1-1][c2]+G[r1-1][c1-1];//区域内的樱桃数量}int solve(int r1,int c1,int r2,int c2){    int& ans=dp[r1][c1][r2][c2];    if(ans!=INF)return ans;    if(count_num(r1,c1,r2,c2)==1)return ans=0;    for(int i=r1;i<r2;i++){        int num1=count_num(r1,c1,i,c2);        int num2=count_num(i+1,c1,r2,c2);//横切        if(num1==0||num2==0)continue;        ans=min(ans,solve(r1,c1,i,c2)+solve(i+1,c1,r2,c2)+c2-c1+1);    }    for(int i=c1;i<c2;i++){        int num1=count_num(r1,c1,r2,i);        int num2=count_num(r1,i+1,r2,c2);//竖切        if(num1==0||num2==0)continue;        ans=min(ans,solve(r1,c1,r2,i)+solve(r1,i+1,r2,c2)+r2-r1+1);    }    return ans;}int main(){    int kase=0;    while(scanf("%d%d%d",&n,&m,&k)==3){        if(k == 0) {            printf("Case %d: %d\n", kase++, 0);        }        memset(G,0,sizeof(G));        int x,y;        for(int i=0;i<k;i++){            scanf("%d%d",&x,&y);            G[x][y]=1;        }        memset(dp,INF,sizeof(dp));        Init();        printf("Case %d: %d\n",++kase,solve(1,1,n,m));    }    return 0;}

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