HDU 4778 Gems Fight!
题意：有n个背包，包里有一些宝石，如今爱丽丝和你轮流选背包，把包里宝石丢到锅中，然后假设锅中有宝石数量到s个，就会得到魔法石，而且能够继续选背包，两人都按最优策略去取，问最后两人魔法石会差多少。
思路：dp,dp[s]表示选背包状态为s时候的值，然后去记忆化搜索就可以，注意假设当前生成魔法石就继续加，否则就减就可以
代码：
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 21;
int g, b, s, dp[(1<<N) + 5], vis[(1<<N) + 5];
int yu[(1<<N) + 5][10];
struct Page {
int num;
int a[10];
} p[N];
int cal(int S, int u) {
int ans = 0;
for (int i = 1; i <= g; i++) {
ans += (yu[S][i] + p[u].a[i]) / s;
}
return ans;
}
int dfs(int now) {
int &tmp = dp[now];
if (vis[now]) return tmp;
vis[now] = 1;
tmp = 0;
int Min = INF, Max = -INF;
if (now == (1<<b) - 1) return tmp;
for (int i = 0; i < b; i++) {
if (now&(1<<i)) continue;
int sb = cal(now, i);
int next = (now|(1<<i));
if (sb != 0) {
Max = max(Max, dfs(next) + sb);
}
else {
Min = min(Min, dfs(next));
}
}
tmp = max(Max, -Min);
return tmp;
}
int main() {
while (~scanf("%d%d%d", &g, &b, &s) && g || b || s) {
for (int i = 0; i < b; i++) {
memset(p[i].a, 0, sizeof(p[i].a));
scanf("%d", &p[i].num);
int tmp;
for (int j = 0; j < p[i].num; j++) {
scanf("%d", &tmp);
p[i].a[tmp]++;
}
}
for (int i = 0; i < (1<<b); i++) {
vis[i] = 0;
for (int j = 0; j < b; j++) {
if (i&(1<<j)) continue;
for (int k = 1; k <= g; k++) {
yu[i|(1<<j)][k] = (yu[i][k] + p[j].a[k]) % s;
}
}
}
printf("%d
", dfs(0));
}
return 0;
}