Uva42417-Headmaster's headache (01 backpack)

Question: uva31617-Headmaster's headache (01 backpack)

This school opens s door tree, giving the principal's existing teachers (n), and then giving m candidates, both the instructors and applicants give the money they need to hire and the subjects they will teach. Make sure that at least two teachers are required to teach each course, and then select a candidate from the course, which requires the minimum cost of employment. Note that the previous teachers should also be included.

Solution: 01 backpack. If every candidate is not hired or will not be hired, then there will be 8 courses, which are represented by 16 persons. If there is a teacher in Class I, put 1 on the binary I-bit. If there are two teachers, put 1 on the I + S.

For example, if there is only one teacher in each of the four courses: 00001111 if there are two teachers, 11110000. Pay attention to the details.

Code:

#include <cstdio>#include <cstring>const int INF = 0x3f3f3f3f;const int N = 105;const int M = 8;const int maxn = 1<<(2 * M);int s, m, n;int T[N][M];int num[N];int cost[N];int dp[maxn][N];int v[M + 1];int mm, sum;int Min (const int a, const int b) { return a < b? a: b;}void init (int st) {for (int i = st; i < mm; i++)for (int j = 1; j <= n + 1; j++)dp[i][j] = INF;for (int i = 1; i <= n + 1; i++)dp[mm][i] = 0;}int DP (int st, int k) {int& ans = dp[st][k];if (ans != INF)return ans;if (k <= n) {int newst = st;for (int j = 0; j < num[k]; j++) {if (newst & (1 << (T[k][j] - 1))) {newst |= (1 << (T[k][j] - 1 + s));newst &= ~(1 << (T[k][j] - 1));} else if (!(newst & (1 << (s + T[k][j] - 1))))newst |= (1 << (T[k][j] - 1)); }//printf ("%d %d\n", st , newst);ans = Min (ans, Min (DP(newst, k + 1) + cost[k], DP(st, k + 1)));}if (ans == INF)ans = INF + 1;return ans;}int main () {int st;int money, t;char ch;while (scanf ("%d%d%d", &s, &m, &n), s || m || n) {st = sum = 0;memset (v, 0, sizeof (v));for (int i = 1; i <= m; i++) {scanf ("%d", &money);sum += money;while (scanf ("%c", &ch)) {if (ch == '\n')break;if (ch >= '1' && ch <= '8') {if (v[ch - '0'] < 2) {v[ch - '0']++;}}}}for (int i = 1; i <= n; i++) {scanf ("%d", &cost[i]);t = 0;while (scanf ("%c", &ch)) {if (ch == '\n')break;if (ch >= '1' && ch <= '8') {T[i][t++] = ch - '0';}}num[i] = t;}for (int i = 1; i <= s; i++)if (v[i])st += 1 << (i - 1 + (v[i] - 1) * s);mm = (1<<(2 * s)) - (1<<s);init (st);printf ("%d\n", DP(st, 1) + sum);}return 0;}