UVa 10032-tug of War

Title: There are n individuals divided into two groups, the difference between the two groups can not exceed 1, find the two groups of people the minimum weight difference.

Analysis: DP, State compression 01 backpack. zoj1880 upgrade version.

First, consider a two-dimensional backpack.

Since it must be placed in a group of two groups, the direct backpack all can go to the state, remove the difference of not more than 1 of the nearest SUM/2 value can be.

Then, optimize.

If the two-dimensional backpack is used directly, due to the large number of data groups will tle, here the use of state-compressed one-dimensional backpack;

Status: F (i) means all possible persons at the time of the total weight of I, the value of f (i), the number on the K-position is 1 represents a combination of K-people.

Transfer: F (i) = f (i) | (f (i-w[j]) << 1), equal to the number of States that can be updated to this state the combination of 1 and the original set.

Finally, scan.

Start scanning down from SUM/2 to find the first solution that satisfies test instructions.

Note: Be aware of the situation of entering 0 persons.

#include <stdio.h> #include <stdlib.h> #include <string.h> long long f[23001];int h[101];int gcd (int A, int b) {return a%b?gcd (b, a%b): b;}    int main () {int t,n,sum,r;        while (scanf ("%d", &t)! = EOF) while (T--) {scanf ("%d", &n);        sum = 0;            for (int i = 1; I <= n; + + i) {scanf ("%d", &h[i]);        Sum + = H[i];        }//wing volume optimization r = 1;        if (n > 0) {r = h[1];        for (int i = 2; I <= n; + + i) R = gcd (r, H[i]);        Sum/= R;                for (int i = 1; I <= n; + + i) h[i]/= R;}         for (int i = SUM/2; I >= 0;--i) f[i] = 0LL;        F[0] = 1LL; for (int i = 1; I <= n; + + i) for (int j = SUM/2; J >= H[i];--j) F[j] |= f[j-h[i]]<<1                ;        int move = sum/2+1;            while (move--) {if (n%2 = = 0 && f[move]& (1ll<< ((n+1)/2)) break; if (n%2 = = 1 && F[move]& (1ll<< ((n+0)/2)) break;                if (n%2 = = 1 && f[move]& (1ll<< ((n+1)/2)) break;}        printf ("%d%d\n", Move*r, (Sum-move) *r);    if (t) printf ("\ n"); } return 0;}

UVa 10032-tug of War