POJ1015 and uva323 Dynamic planning 01 Backpack Extensions

Solution: At first thought this accord with the optimal sub-structure of the dynamic planning, but later looked at other people's solution found that does not conform to the optimal sub-structure, can not transfer all the state. Because maybe I picked the front one. But if I choose the other one plus the better, this time chose not to choose another because it has disappeared, this time can be done with 01 backpack and a vector array to record the path, with the array of the second dimension 20* M is an absolute difference of 2 sides representing an absolute difference of 1.2 ... 3 ...

This allows each state to be transferred past.

#include <iostream> #include <cmath> #include <cstdio> #include <algorithm> #include <
Cstring> #include <vector> using namespace std;
const int MX = 205;
const int inf = 0X3F3F3F3F;
int DP[MX][4*MX];
vector<int>pre[mx][4*mx];
int DE[MX],SU[MX];
    int main () {int n,m;
    int casei = 1; while (scanf ("%d%d", &n,&m), n| |
        m) {int A, B;
        for (int i = 1, i <= m; i++) for (int j = 1; j < 4*mx; J + +) Pre[i][j].clear ();  for (int i = 1; I <= N;
            i++) {scanf ("%d%d", &a,&b);
            De[i] = a-B;
        Su[i] = a+b;
        } memset (Dp,-1,sizeof (DP));
        int f = 20*m;
        DP[0][F] = 0; for (int i = 1, i <= N; i++) for (int j = m-1; J >= 0; j--) for (int k = 0; k < 2*f; k++ ) if (dp[j][k]!=-1&& (k+de[i]>=0&&k+de[i]<=2*f)) {if (dp[j+ 1][k+de[i]] < dp[J][k]+su[i]) {Dp[j+1][k+de[i]] = dp[j][k]+su[i];
                            Pre[j+1][k+de[i]] = pre[j][k];
                        Pre[j+1][k+de[i]].push_back (i);
        }} int i;
        for (i = 0; Dp[m][f+i] = = -1&&dp[m][f-i] = =-1; i++);
        int temp = dp[m][f+i]>dp[m][f-i]?f+i:f-i;
        int sum1 = (dp[m][temp]+temp-f)/2;
        int sum2 = (dp[m][temp]-temp+f)/2;
        printf ("Jury #%d\n", casei++);
        printf ("Best jury have value%d for prosecution and value%d for defence:\n", sum1,sum2);
        for (int i = 0; i < m; i++) printf ("%d", pre[m][temp][i]);
    printf ("\ n");
} return 0;
 }