演算法作業三_0-1 Double Knapsack Problem_4

#define _CRT_SCURE_NO_WARNINGS
#include <iostream>

int main() {
	int N = 0, W = 0, V = 0;
	int c[100] = { 0 }, w[100] = { 0 }, v[100] = { 0 };
	int s[100] = { 0 };
	int pw = 0, pv = 0, T = 0, q = 0;
	int ans[100] = { 0 };
	static int t[101][1001][1001] = { 0 };
	scanf("%d", &N);
	scanf("%d", &W);
	pw = W;
	scanf("%d", &V);
	pv = V;
	for (int i = 1; i <= N; i++) {
		scanf("%d", &c[i]);
	}
	for (int i = 1; i <= N; i++) {
		scanf("%d", &w[i]);
	}
	for (int i = 1; i <= N; i++) {
		scanf("%d", &v[i]);
	}

	for (int i = 1; i <= N; i++) {
		for (int j = 1; j <= W; j++) {
			for (int k = 1; k <= V; k++) {
				if (w[i] > j || v[i] > k) {
					t[i][j][k] = t[i - 1][j][k];
				}
				else {
					if (t[i - 1][j][k] > c[i] + t[i - 1][j - w[i]][k - v[i]]) {
						t[i][j][k] = t[i - 1][j][k];
					}
					else {
						t[i][j][k] = c[i] + t[i - 1][j - w[i]][k - v[i]];
					}
				}
			}
		}
	}

	int text = 0;
	for (int i = N; i >= 1; i--) {
		if (t[i][pw][pv] == t[i - 1][pw][pv]) {
			s[i] = 0;
		}
		else {
			s[i] = 1;
			T = T + c[i];
			q = q + 1;
			pw = pw - w[i];
			pv = pv - v[i];
			ans[text] = i;
			text = text + 1;
		}
	}

	for (int i = 0; i < text-1; i++) {
		for (int j = 0; j < text - 1; j++) {
			if (ans[j] > ans[j + 1]) {
				int temp = 0;
				temp = ans[j];
				ans[j] = ans[j + 1];
				ans[j + 1] = temp;
			}
		}
	}


	printf("%d\n%d\n(", T, q);
	printf("%d",ans[0]);
	for (int i = 1; i < text; i++) {
		printf(",%d", ans[i]);
	}
	printf(")");
}