hdu 4778 Rabbit Kingdom(减少国家)

题目链接：hdu 4778 Rabbit Kingdom

题目大意：Alice和Bob玩游戏，有一个炉子。能够将S个同样颜色的宝石换成一个魔法石。如今有B个包，每一个包里有若干个宝石，给出宝石的颜色。如今由Alice開始，两人轮流选取一个包的宝石放入炉中，每当获得一个魔法石时，能够额外获得一次机会再选一个包放入。两人均依照自己的最优策略。问说最后Alice的魔法石-Bob的魔法石是多少。

解题思路：状态压缩，221,对于每次移动到下一个状态，假设获得的魔法石g非零。则说明下一个状态还是自己在取。则要选择最优的。假设g为0。则说明下一个状态不是自己在取，则要取尽量小的，相应也就是相反数尽量大的。

C++ 记忆化版
 
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxb = 21;
const int maxn = 1<<21;
const int maxg = 10;
const int INF = 0x3f3f3f3f;
int G, B, S;
int c[maxb+5][maxg], s[maxn+5][maxg];
int v[maxn+5], dp[maxn+5];
void init () {
    int t, a;
    memset(c, 0, sizeof(c));
    memset(v, 0, sizeof(v));
    for (int i = 0; i < B; i++) {
        scanf("%d", &t);
        for (int j = 0; j < t; j++) {
            scanf("%d", &a);
            c[i][a]++;
        }
    }
    for (int i = 0; i < (1<<B); i++) {
        for (int j = 0; j < B; j++) {
            if (i&(1<<j))
                continue;
            int e = i|(1<<j);
            if (v[e])
                continue;
            for (int k = 1; k <= G; k++)
                s[e][k] = (s[i][k] + c[j][k]) % S;
        }
    }
}
int add (int k, int x) {
    int ans = 0;
    for (int i = 1; i <= G; i++)
        ans += (s[k][i] + c[x][i]) / S;
    return ans;
}
int dfs (int u) {
    if (u + 1 == (1<<B))
        return 0;
    if (v[u])
        return dp[u];
    int up = -INF;
    int lower = INF;
    for (int i = 0; i < B; i++) {
        if (u&(1<<i))
            continue;
        int g = add (u, i);
        if (g)
            up = max(up, dfs(u|(1<<i))+g);
        else
            lower = min(lower, dfs(u|1<<i));
    }
    v[u] = 1;
    return dp[u] = max(up, -lower);
}
int main () {
    while (scanf("%d%d%d", &G, &B, &S) == 3 && G + B + S) {
        init();
        memset(v, 0, sizeof(v));
        printf("%d\n", dfs(0));
    }
    return 0;
}

C++ 递推版
 
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxb = 21;
const int maxn = 1<<21;
const int maxg = 10;
const int INF = 0x3f3f3f3f;
int G, B, S;
int c[maxb+5][maxg], s[maxn+5][maxg];
int v[maxn+5], dp[maxn+5];
void init () {
    int t, a;
    memset(c, 0, sizeof(c));
    memset(v, 0, sizeof(v));
    for (int i = 0; i < B; i++) {
        scanf("%d", &t);
        for (int j = 0; j < t; j++) {
            scanf("%d", &a);
            c[i][a]++;
        }
    }
    for (int i = 0; i < (1<<B); i++) {
        for (int j = 0; j < B; j++) {
            if (i&(1<<j))
                continue;
            int e = i|(1<<j);
            if (v[e])
                continue;
            for (int k = 1; k <= G; k++)
                s[e][k] = (s[i][k] + c[j][k]) % S;
        }
    }
}
int add (int k, int x) {
    int ans = 0;
    for (int i = 1; i <= G; i++)
        ans += (s[k][i] + c[x][i]) / S;
    return ans;
}
int solve () {
    int e = (1<<B) - 1;
    memset(dp, -INF, sizeof(dp));
    dp[e] = 0;
    for (int u = e; u >= 0; u--) {
        for (int i = 0; i < B; i++) {
            if ((u&(1<<i)) == 0)
                continue;
            int ui = u-(1<<i);
            int g = add(ui, i);
            if (g)
                dp[ui] = max(dp[ui], dp[u] + g);
            else
                dp[ui] = max(dp[ui], -dp[u]);
        }
    }
    return dp[0];
}
int main () {
    while (scanf("%d%d%d", &G, &B, &S) == 3 && G + B + S) {
        init();
        printf("%d\n", solve());
    }
    return 0;
}

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