Building Block Castle (Castle) (01 backpack)

Castle)

Time limit:

1000 ms

Memory limit:

102400kb

Description

There are N sequences that are not incrementing. Now we need to remove some of them to make the sum of each sequence as big as possible, and this sum should be as big as possible.

Input

The first line is an integer n (n <= 100), indicating a total of several castles. Each line in the N rows below is a series of non-negative integers separated by a space, and the edges of all blocks in a castle are given in sequence from bottom to top. End with-1. The number of blocks in a castle cannot exceed 100, and the length of each block cannot exceed 100.

Output

An integer that represents the maximum possible height of the final Castle. If no suitable solution is found, 0 is output.

Sample Input

2

2 1-1

3 2 1-1

Sample output

3

Question: Use a 01 backpack for each castle to find all the heights that meet the conditions. Then count the highest heights of all the castles.

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;int h[101];bool dp[101][10001];int main(){    int n, ret;    while(scanf("%d",&n) != EOF)    {        ret = 10000;        int i, j, k, cnt, sum;        for(k = 0; k < n; k++)        {            cnt = sum = 0;            while(1)            {                scanf("%d",&h[cnt]);                if(h[cnt] == -1) break;                sum += h[cnt++];            }            dp[k][0] = true;            for(i = 0; i < cnt; i++)                for(j = sum; j >= h[i]; j--)                    dp[k][j] = (dp[k][j] || dp[k][j-h[i]]);            for(i = sum; !dp[k][i]; i--);            ret = min(ret, i);        }        for(i = ret; i >= 0; i--)        {            bool flag = true;            for(j = 0; j < n; j++)                if(dp[j][i] == false) {flag = false; break; }            if(flag == true) { ret = i; break;}        }        printf("%d\n",ret);    }    return 0;}