POJ 多重背包专题

最新推荐文章于 2021-08-26 12:59:36 发布

最新推荐文章于 2021-08-26 12:59:36 发布 · 170 阅读

POJ 1014 Dividing

这道题用背包做有两种解法，一种是拆分法，另一种是很神的O(VN)的DP法。

拆分法：

#include <iostream> #include <vector> #include <list> #include <map> #include <set> #include <deque> #include <queue> #include <stack> #include <bitset> #include <algorithm> #include <functional> #include <numeric> #include <utility> #include <sstream> #include <iomanip> #include <cstdio> #include <cmath> #include <cstdlib> #include <cctype> #include <string> #include <cstring> #include <cmath> #include <ctime> #define MAXN 10005 #define INF 100000000 #define eps 1e-11 #define L(x) x<<1 #define R(x) x<<1|1 using namespace std; int f[240005], v; void complete_pack(int *a, int c, int w) { for(int i = c; i <= v; i++) a[i] = max(a[i], a[i - c] + w); } void zeroone_pack(int *a, int c, int w) { for(int i = v; i >= c; i--) a[i] = max(a[i], a[i - c] + w); } void mutiple_pack(int *a, int c, int w, int M) { if(c * M >= v) { complete_pack(a, c, w); return; } int k = 1; while(k < M) { zeroone_pack(a, k * c, k * w); M = M - k; k = 2 * k; } zeroone_pack(a, c * M, w * M); } int main() { //freopen("d:/data.in","r",stdin); //freopen("d:/data.out","w",stdout); int sum, i, c[7], w[7], m[7],cas = 0; while(scanf("%d%d%d%d%d%d", &m[1], &m[2], &m[3], &m[4], &m[5], &m[6])) { if(m[1] == 0 && m[2] == 0 && m[3] == 0 && m[4] == 0 && m[5] == 0 && m[6] == 0) break; sum = 0; for(i = 1; i <= 6; i++) { c[i] = w[i] = i; sum += c[i] * m[i]; } printf("Collection #%d:\n", ++cas); if(sum & 1) { puts("Can't be divided.\n"); } else { sum /= 2; for(i = 1; i <= sum; i++) f[i] = -INF; f[0] = 0; v = sum; for(i = 1; i <= 6; i++) mutiple_pack(f, c[i], w[i], m[i]); if(f[v] < 0) { puts("Can't be divided.\n"); } else { puts("Can be divided.\n"); } } } return 0; }

然后是O（VN）的方法，因为本题是一种背包的可行性问题，所以才能用这种方法。

#include <iostream> #include <vector> #include <list> #include <map> #include <set> #include <deque> #include <queue> #include <stack> #include <bitset> #include <algorithm> #include <functional> #include <numeric> #include <utility> #include <sstream> #include <iomanip> #include <cstdio> #include <cmath> #include <cstdlib> #include <cctype> #include <string> #include <cstring> #include <cmath> #include <ctime> #define MAXN 10005 #define INF 100000000 #define eps 1e-11 #define L(x) x<<1 #define R(x) x<<1|1 using namespace std; int f[240005], v, used[240005]; int main() { //freopen("d:/data.in","r",stdin); //freopen("d:/data.out","w",stdout); int sum, i, c[7], w[7], m[7],cas = 0, j; while(scanf("%d%d%d%d%d%d", &m[1], &m[2], &m[3], &m[4], &m[5], &m[6])) { if(m[1] == 0 && m[2] == 0 && m[3] == 0 && m[4] == 0 && m[5] == 0 && m[6] == 0) break; sum = 0; for(i = 1; i <= 6; i++) { c[i] = w[i] = i; sum += c[i] * m[i]; } printf("Collection #%d:\n", ++cas); if(sum & 1) { puts("Can't be divided.\n"); } else { sum /= 2; memset(f, 0, sizeof(f)); f[0] = 1; for(i = 1; i <= 6; i++) { for(j = 0; j <= sum; j++) used[j] = 0; for(j = c[i]; j <= sum; j++) { if(!f[j] && f[j - c[i]] && used[j - c[i]] + 1 <= m[i]) { f[j] = 1; used[j] = used[j - c[i]] + 1; } } } if(f[sum]) { puts("Can be divided.\n"); } else puts("Can't be divided.\n"); } } return 0; }

POJ 1742Coins

很裸的背包可行性问题

POJ 2392 Space Elevator

这个题的多了一个限制条件，实际上只要按限制条件从小到大排序就行

#include <iostream> #include <vector> #include <list> #include <map> #include <set> #include <deque> #include <queue> #include <stack> #include <bitset> #include <algorithm> #include <functional> #include <numeric> #include <utility> #include <sstream> #include <iomanip> #include <cstdio> #include <cmath> #include <cstdlib> #include <cctype> #include <string> #include <cstring> #include <cmath> #include <ctime> #define MAXN 10005 #define INF 100000000 #define eps 1e-11 #define L(x) x<<1 #define R(x) x<<1|1 using namespace std; struct wwj { int h, num, a; }p[405]; bool f[400005]; int used[400005]; bool cmp(wwj x, wwj y) { return x.a < y.a; } int main() { //freopen("d:/data.in","r",stdin); //freopen("d:/data.out","w",stdout); int k, i, j; scanf("%d", &k); for(i = 1; i <= k; i++) { scanf("%d%d%d", &p[i].h, &p[i].a, &p[i].num); } sort(p + 1, p + k + 1, cmp); memset(f, 0, sizeof(f)); f[0] = 1; int ans = 0; for(i = 1; i <= k; i++) { memset(used, 0, sizeof(used)); for(j = p[i].h; j <= p[i].a; j++) { if(!f[j] && f[j - p[i].h] && used[j - p[i].h] + 1 <= p[i].num) { f[j] = 1; used[j] = used[j - p[i].h] + 1; ans = max(j, ans); } } } printf("%d\n", ans); return 0; }
POJ 1276Cash Machine

裸题不解释

其实就是个0-1背包变形，对同种颜色的衣服，把一个人洗所有衣服的时间算出来，除以二，然后看能达到的最大容量，这就可以保证两人的洗衣时间尽量平均了。也可以用多重背包的做法做，只要把件数都设成1就行。

#include <iostream> #include <vector> #include <list> #include <map> #include <set> #include <deque> #include <queue> #include <stack> #include <bitset> #include <algorithm> #include <functional> #include <numeric> #include <utility> #include <sstream> #include <iomanip> #include <cstdio> #include <cmath> #include <cstdlib> #include <cctype> #include <string> #include <cstring> #include <cmath> #include <ctime> #define MAXN 10005 #define INF 100000000 #define eps 1e-11 #define L(x) x<<1 #define R(x) x<<1|1 using namespace std; bool f[50005]; int v[11][101], cnt[11]; int main() { //freopen("d:/data.in","r",stdin); //freopen("d:/data.out","w",stdout); int n, m, i, j, x, k; char s[22]; while(scanf("%d%d", &n, &m) != EOF) { if(n == 0 && m == 0) break; map<string, int>mp; memset(cnt, 0, sizeof(cnt)); for(i = 1; i <= n; i++) { scanf("%s", s); string tmp = s; mp[tmp] = i; } for(i = 1; i <= m; i++) { scanf("%d%s", &x, s); string tmp = s; cnt[mp[tmp]]++; v[mp[tmp]][cnt[mp[tmp]]] = x; } int ans = 0; int mx; for(i = 1; i <= n; i++) { int sum = 0; mx = 0; for(j = 1; j <= cnt[i]; j++) sum += v[i][j]; for(j = 0; j <= sum / 2; j++) f[j] = 0; f[0] = 1; for(j = 1; j <= cnt[i]; j++) { for(k = sum / 2; k >= v[i][j]; k--) //倒序的0-1背包做法 { if(!f[k] && f[k - v[i][j]] ) { f[k] = 1; mx = max(mx, k); } } /* 多重背包做法 for(k = 0; k <= sum / 2; k++) used[k] = 0; for(k = v[i][j]; k <= sum / 2; k++) { if(!f[k] && f[k - v[i][j]] && !used[k - v[i][j]]) { used[k] = 1; f[k] = 1; mx = max(mx, k); } } */ } ans += (sum - mx); } printf("%d\n", ans); } return 0; }