狀態壓縮·一
題目傳送:#1044 : 狀態壓縮·一
AC代碼:
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <complex>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <sstream>
#include <utility>
#include <iostream>
#include <algorithm>
#include <functional>
#define LL long long
#define INF 0x7fffffff
using namespace std;
int n, m, q;
int w[1005];
int dp[1005][1030];//dp[i][j]表示選到第i個位置時j狀態能夠取得的最大值
int cnt[1030];//代表每個數的位數上的1的個數
int main() {
cnt[0] = 0, cnt[1] = 1;
for(int i = 2; i < 1030; i ++) cnt[i] = cnt[i >> 1] + cnt[i & 1];
scanf("%d %d %d", &n, &m, &q);
for(int i = 1; i <= n; i ++) {
scanf("%d", &w[i]);
}
int ans = 0;
int d = 1 << m;
for(int i = 1; i <= n; i ++) {
for(int j = 0; j < (1 << m); j ++) {
if(cnt[j] <= q) dp[i][j] = max(dp[i-1][j >> 1], dp[i-1][(j >> 1) + (1 << (m - 1))]) + (j & 1) * w[i];
ans = max(ans, dp[i][j]);
}
}
printf("%d\n", ans);
return 0;
}
Hackers’ Crackdown
題目傳送:UVA - 11825 - Hackers’ Crackdown
AC代碼:
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <complex>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <sstream>
#include <utility>
#include <iostream>
#include <algorithm>
#include <functional>
#define LL long long
#define INF 0x7fffffff
using namespace std;
const int maxn = (1 << 16) + 5;
int dp[maxn];//dp[i]表示子集i最多可以分成多少組
int cover[maxn];//cover[i]表示若干集合(i中所表示的集合)的並集
int n, m;
int P[25];//P[i]表示與i相連的數的集合(包括i)
int main() {
int cas = 1;
while(scanf("%d", &n) != EOF) {
if(n == 0) break;
for(int i = 0; i < n; i ++) {
scanf("%d", &m);
P[i] = 1 << i;
int t;
while(m --) {
scanf("%d", &t);
P[i] |= (1 << t);//併入集合i
}
}
for(int i = 0; i < maxn; i ++) {
cover[i] = 0;
for(int j = 0; j < n; j ++) {//if判斷j是否在i中,是得話就併入
if(i & (1 << j)) cover[i] |= P[j];
}
}
dp[0] = 0;
int tot = (1 << n) - 1;//全集總數
for(int i = 1; i <= tot; i ++) {//依次枚舉全集,因爲要先算出前一狀態才能推出後一狀態
dp[i] = 0;
for(int j = i; j; j = (j - 1) & i) {//枚舉子集的技巧,重點!
if(cover[j] == tot) {//i的子集j等於全集,則執行狀態轉移
dp[i] = max(dp[i], dp[i ^ j] + 1);
}
}
}
printf("Case %d: %d\n", cas ++, dp[tot]);
}
return 0;
}
Sharing Chocolate
題目傳送:UVALive - 4794 - Sharing Chocolate
WF2010的題。
AC代碼:
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <complex>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <sstream>
#include <utility>
#include <iostream>
#include <algorithm>
#include <functional>
#define LL long long
#define INF 0x7fffffff
using namespace std;
const int maxn = 16;
int d[1 << maxn][105];
int vis[1 << maxn][105];
int sum[1 << maxn];
int a[maxn];
int n;
int bitcount(int x) {
return x == 0 ? 0 : bitcount(x / 2) + (x & 1);
}
int dp(int s, int x) {//每次遞歸找出集合爲s,寬爲x的巧克力是否可以滿足要求
if(vis[s][x]) return d[s][x];
vis[s][x] = 1;
int& ans = d[s][x];
if(bitcount(s) == 1) return ans = 1;//此時爲邊界,即只有一塊巧克力的情況,肯定是滿足的
int y = sum[s] / x;//另一個邊長可以根據這個求得
for(int s0 = (s - 1) & s; s0; s0 = (s0 - 1) & s) {//枚舉子集
int s1 = s - s0;
if(sum[s0] % x == 0 && dp(s0, min(x, sum[s0] / x)) && dp(s1, min(x, sum[s1] / x))) return ans = 1;//豎着切(這裏假定寬x是豎着的)
if(sum[s0] % y == 0 && dp(s0, min(y, sum[s0] / y)) && dp(s1, min(y, sum[s1] / y))) return ans = 1;//橫着切
}
return ans = 0;
}
int main() {
int cas = 1, x, y;
while(scanf("%d", &n) != EOF) {
if(n == 0) break;
scanf("%d %d", &x, &y);
for(int i = 0; i < n; i ++) scanf("%d", &a[i]);
//計算每個子集的元素的和
memset(sum, 0, sizeof(sum));
for(int s = 0; s < (1 << n); s ++) {
for(int i = 0; i < n; i ++) if(s & (1 << i)) sum[s] += a[i];
}
memset(vis, 0, sizeof(vis));
int ALL = (1 << n) - 1;
int ans;
if(sum[ALL] != x * y) ans = 0;
else ans = dp(ALL, min(x, y));
printf("Case %d: %s\n", cas ++, ans ? "Yes" : "No");
}
return 0;
}