HDU 6006 Engineer Assignment (状压DP)

题意：一共有n个任务，完成某个任务需要会一些领域的人，一共有m个工程师，每个工程师会一些领域，问这些工程师最多完成多少任务。

析：一个简单的状压DP，在比赛，算着时间复杂度过不了，结果才15ms，说一下思路，先预处理每个工程能有哪几种工程师来完成，然后dp[i][s] 表示前 i 个任务，工程师状态为要来完成的最多几个工程，dp[i][s] = max { dp[j][s^x] + 1 }。其中 x 是完成 i 工程所以需要的工程师。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
//#define mp make_pair
#define cl clear()
//#define all 1,n,1
#define FOR(x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e18;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 100;
const LL mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r > 0 && r <= n && c > 0 && c <= m;
}

int dp[15][1500];
int pst[20], est[20];
int plen[20], elen[20];
int cnt;
map<int, int> mp;

int ID(int x){
if(mp.count(x))  return mp[x];
return mp[x] = cnt++;
}

vector<int> v[15];

int main(){
int T;  cin >> T;
for(int kase = 1; kase <= T; ++kase){
scanf("%d %d", &n, &m);
mp.cl

;  cnt = 0;
for(int i = 0; i < n; ++i){
scanf("%d", &plen[i]);
pst[i] = 0;
for(int j = 0; j < plen[i]; ++j){
int y;
scanf("%d", &y);
pst[i] |= 1<<ID(y);
}
}

for(int i = 0; i < m; ++i){
scanf("%d", &elen[i]);
est[i] = 0;
for(int j = 0; j < elen[i]; ++j){
int y;
scanf("%d", &y);
if(mp.count(y))  est[i] |= 1<<mp[y];
}
}

for(int i = 0; i < n; ++i){
int st = 0;
v[i].cl;
for(int j = 0; j < m; ++j)
if((pst[i] & est[j]) == pst[i])  v[i].pb(1<<j);
if(plen[i] > 1){
for(int j = 0; j < m; ++j)
for(int k = j+1; k < m; ++k)
if((pst[i] & (est[j]|est[k])) == pst[i])  v[i].pb(1<<j|1<<k);
}
if(plen[i] > 2){
for(int j = 0; j < m; ++j)  for(int k = j+1; k < m; ++k)
for(int l = k+1; l < m; ++l)
if((pst[i] & (est[j]|est[k]|est[l])) == pst[i])  v[i].pb(1<<j|1<<k|1<<l);
}
}

ms(dp, 0);
int ans = 0;
for(int i = 0; i < v[0].sz; ++i){
dp[0][v[0][i]] = 1;
ans = max(ans, dp[0][v[0][i]]);
}
int all = 1<<m;
for(int i = 1; i < n; ++i){
for(int j = 1; j < all; ++j){
for(int k = 0; k < v[i].sz; ++k){
if((v[i][k]&j) != v[i][k])  continue;
for(int l = 0; l < i; ++l)
dp[i][j] = max(dp[i][j], dp[l][j^v[i][k]]+1);
ans = max(ans, dp[i][j]);
}
}
}

printf("Case #%d: %d\n", kase, ans);
}
return 0;
}