Engineer Assignment HDU - 6006 状压dp

http://acm.split.hdu.edu.cn/showproblem.php?pid=6006

比赛的时候写了一个暴力，存暴力，过了，还46ms

那个暴力的思路是，预处理can[i][j]表示第i个人能否胜任第j个项目，能胜任的条件就是它和这个项目有共同的需求。

然后暴力枚举每一个人去搭配哪一个项目，

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <string>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
#define X first
#define Y second
#define clr(u,v); memset(u,v,sizeof(u));
#define inff() freopen("data","r",stdin);
#define out() freopen("ans","w",stdout);
#define Clear(Q); while (!Q.empty()) Q.pop();
#define pb push_back
using namespace std;
typedef long long ll;
typedef long long LL;
typedef pair <int, int> pii;
const LL INF = 1e17;
const int inf = 0x3f3f3f3f;
const int maxn = 1e2 + 2;
vector<int> project[maxn], man[maxn];
vector<int> state[maxn];
int solve[12][maxn], DFN;
int dp[12][1024 + 2];
void work() {
int n, m; // n_project, m_man
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; ++i) { // project
int c;
scanf("%d", &c);
project[i].clear();
while (c--) {
int val;
scanf("%d", &val);
project[i].push_back(val);
}
}
for (int i = 1; i <= m; ++i) { // man
int c;
scanf("%d", &c);
man[i].clear();
while (c--) {
int val;
scanf("%d", &val);
man[i].push_back(val);
}
}
int en = (1 << m) - 1;
for (int i = 1; i <= n; ++i) {
state[i].clear();
for (int j = 1; j <= en; ++j) {
++DFN;
for (int k = 1; k <= m; ++k) {
if (j & (1 << (k - 1))) {
for (int d = 0; d < man[k].size(); ++d) {
solve[i][man[k][d]] = DFN;
}
}
}
bool flag = true;
for (int d = 0; d < project[i].size(); ++d) {
if (solve[i][project[i][d]] != DFN) {
flag = false;
break;
}
}
if (flag) state[i].push_back(j);
}
}
//    for (int i = 1; i <= n; ++i) {
//        for (int j = 0; j < state[i].size(); ++j) {
//            printf("%d ", state[i][j]);
//        }
//        printf("\n");
//    }
memset(dp, false, sizeof dp);
for (int i = 1; i <= n; ++i) {
for (int d = 1; d <= en; ++d) {
dp[i][d] = dp[i - 1][d]; // 不做这个项目
for (int k = 0; k < state[i].size(); ++k) {
if ((d | state[i][k]) > d) continue;
int res = d ^ state[i][k];
dp[i][d] = max(dp[i][d], dp[i - 1][res] + 1);
}
}
}
int ans = 0;
for (int i = 1; i <= en; ++i) ans = max(ans, dp
[i]);
static int f = 0;
printf("Case #%d: %d\n", ++f, ans);
}

int main() {
#ifdef local
freopen("data.txt", "r", stdin);
#endif
int T;
scanf("%d", &T);
while(T--) {
work();
}
return 0;
}

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