POJ1837Balance(多重背包计数)-电脑资料 BalanceTime Limit:1000MSMemory Limit:30000KTotal Submissions:11706Accepted:7305 Description Gigel has a strange balance and he wants to poise it. Actually, the device is different from any other ordinary balance. It orders two arms of negligible weight and ea...
每一个新阶段都是在原来的出现过的状态上更新的。 状态方程dp [ i ][ j+ w[ i ]*c[ k ] ]= ∑(dp[ i-1 ][ j ] ); 下面是代码: #include <stdio.h> #include <string.h> int dp[21][15005]; int main() { int c,g,i,j,k; while(scanf("%d%d",&c,&g)!=EOF) { int c1[...
代码如下: import java.io.BufferedReader; import java.io.File; import java.io.FileInputStream; import java.io.IOException; import java.io.InputStream; import java.io.InputStreamReader; import java.io.PrintWriter; import java.util.Arrays; import java.util.Map; import java.util.StringTokenizer; p...
for(int j = 0; j <= 15000; j ++) { // enum balance value for(int k = 1; k <= n; k ++) { // enum position if(dp[i-1][j] == INF) continue; int newBalance = (pos[k]*weight[i]) + j; if(dp[i][newBalance] == INF) { dp[i][newBalance] = dp[i-1][j]; }...
POJ 1837 Balance(DP) POJ 1837 Balance 题意:给一个天平,天平的左右臂长都为15,现在给c个挂钩位置和g个质量的勾码,求有几种方式放勾码可以使天平保持平衡。 解析:用到DP,所以要先弄清DP中的i和j是干什么用的,i的作用是用于存用了几个勾码,j的作用是用于存状态,比如都放在一边的头上,就...
BalanceTime Limit:1000MSMemory Limit:30000KTotal Submissions:11706Accepted:7305 Description Gigel has a strange balance and he wants to poise it. Actually, the device is different from any other ordinary balance. It orders two arms of negligible weight and each arm's length is 15. Some hooks ...
(1)背包问题. (poj1837,poj1276)(2)型如下表的简单DP(可参考lrj的书 page149):1.E[j]=opt{D...
(1)背包问题. (poj1837,poj1276)(2)型如下表的简单DP(可参考lrj的书 page149):1.E[j]=opt{D...
五.动态规划 (1)背包问题. (poj1837,poj1276) (2)型如下表的简单DP(可参考lrj的书 page149): 1.E[j]=opt{D[i]+w(i,j)} (poj3267,poj1836,poj1260,poj2533) 2.E[i,j]=opt{D[i-1,j]+xi,D[i,j-1]+yj,D[i-1][j-1]+zij} (最长公共子序列) (poj3176,poj1080,poj1159) ...