hdu 4778 Rabbit Kingdom(状态压缩)

题目链接：hdu 4778 Rabbit Kingdom

题目大意：Alice和Bob玩游戏，有一个炉子，可以将S个相同颜色的宝石换成一个魔法石，现在有B个包，每个包里有若干个宝石，给出宝石的颜色。现在由Alice开始，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，可以额外获得一次机会再选一个包放入。两人均按照自己的最优策略，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，如果获得的魔法石g非零，则说明下一个状态还是自己在取，则要选择最优的。如果g为0，则说明下一个状态不是自己在取，则要取尽量小的，对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;
}