标签：noi 之间 分享 i+1 poj sigma include row turn
http://poj.org/problem?id=1191
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 15655 | Accepted: 5556 |
Description
Input
Output
Sample Input
3 1 1 1 1 1 1 1 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 0 3
Sample Output
1.633
Source
/* 设f(i,a,b,c,d)表示切第i刀，剩余的矩形左上角和右下角的坐标是(a,b)和(c,d)， 除了剩余部分其它部分的xi平方和的最小值。 那么f(i)可以向f(i+1)转移，只需要暴力枚举第i+1刀从哪里切了一刀即可。 */ #include <iostream> #include <cstdio> #include <cmath> using namespace std; const int inf=1<<30; int n, chess[9][9],sum[9][9],dp[9][9][9][9][15]; int getX(int y1, int x1, int y2, int x2) { int a=sum[y2][x2]-sum[y2][x1-1]-sum[y1-1][x2]+sum[y1-1][x1-1]; return a*a; } int main() { scanf("%d", &n); for(int i=1; i<=8; i++) for(int j=1; j<=8; j++) scanf("%d", &chess[i][j]); for(int i=1; i<=8; i++) { for(int j=1; j<=8; j++) sum[i][j]=sum[i][j-1]+chess[i][j]; for(int j=1; j<=8; j++) sum[i][j]+=sum[i-1][j]; } for(int i1=1; i1<=8; i1++) for(int j1=1; j1<=8; j1++) for(int i2=i1; i2<=8; i2++) for(int j2=j1; j2<=8; j2++) dp[i1][j1][i2][j2][0]=getX(i1, j1, i2, j2); for(int i=1; i<n; i++) for(int i1=1; i1<=8; i1++) for(int j1=1; j1<=8; j1++) for(int i2=i1; i2<=8; i2++) for(int j2=j1; j2<=8; j2++) { dp[i1][j1][i2][j2][i]=inf; //左右切割 for(int k=j1; k<j2; k++) dp[i1][j1][i2][j2][i]=min(dp[i1][j1][i2][j2][i], min(dp[i1][j1][i2][k][i-1]+dp[i1][k+1][i2][j2][0], dp[i1][j1][i2][k][0]+dp[i1][k+1][i2][j2][i-1])); //上下切割 for(int k=i1; k<i2; k++) dp[i1][j1][i2][j2][i]=min(dp[i1][j1][i2][j2][i], min(dp[i1][j1][k][j2][i-1]+dp[k+1][j1][i2][j2][0], dp[i1][j1][k][j2][0]+dp[k+1][j1][i2][j2][i-1])); } printf("%d\n",dp[1][1][8][8][n-1]); return 0; }
标签：noi 之间 分享 i+1 poj sigma include row turn
原文地址：http://www.cnblogs.com/L-Memory/p/7658019.html
