codeforces 580D(状态dp)
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When Kefa came to the restaurant and sat at a table, the waiter immediately brought him the menu. There were n dishes. Kefa knows that he needs exactly m dishes. But at that, he doesn't want to order the same dish twice to taste as many dishes as possible.
Kefa knows that the i-th dish gives him ai units of satisfaction. But some dishes do not go well together and some dishes go very well together. Kefa set to himself k rules of eating food of the following type — if he eats dish x exactly before dish y (there should be no other dishes between x and y), then his satisfaction level raises by c.
Of course, our parrot wants to get some maximal possible satisfaction from going to the restaurant. Help him in this hard task!
The first line of the input contains three space-separated numbers, n, m and k (1 ≤ m ≤ n ≤ 18, 0 ≤ k ≤ n * (n - 1)) — the number of dishes on the menu, the number of portions Kefa needs to eat to get full and the number of eating rules.
The second line contains n space-separated numbers ai, (0 ≤ ai ≤ 109) — the satisfaction he gets from the i-th dish.
Next k lines contain the rules. The i-th rule is described by the three numbers xi, yi and ci (1 ≤ xi, yi ≤ n, 0 ≤ ci ≤ 109). That means that if you eat dish xi right before dish yi, then the Kefa's satisfaction increases by ci. It is guaranteed that there are no such pairs of indexesi and j (1 ≤ i < j ≤ k), that xi = xj and yi = yj.
In the single line of the output print the maximum satisfaction that Kefa can get from going to the restaurant.
2 2 11 12 1 1
3
4 3 21 2 3 42 1 53 4 2
12
In the first sample it is best to first eat the second dish, then the first one. Then we get one unit of satisfaction for each dish and plus one more for the rule.
In the second test the fitting sequences of choice are 4 2 1 or 2 1 4. In both cases we get satisfaction 7 for dishes and also, if we fulfill rule 1, we get an additional satisfaction 5.
题目分析:
A two-dimensional DP will be used to solve the problem. The first dimention is the mask of already taken dishes, and the second — the number of the last taken dish. We will go through all the zero bits of the current mask for the transitions. We will try to put the one in them, and then update the answer for a new mask. The answer will consist of the answer of the old mask, a dish value, which conforms to the added bit and the rule, that can be used. The final answer is the maximum of all the values of DP, where mask contains exactly mones.
Asymptotics — O(2n * n2).
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int n, m, k;const int maxn = 1048576;long long a[20], dp[maxn][20], b[maxn], c[20][20];int len(int x){ int k = 0; for(int i = 0; i < n; ++i) if(x&b[i]) k++; return k;}int main(){ while(~scanf("%d%d%d", &n, &m, &k)){ for(int i = 0; i < n; ++i) scanf("%I64d", &a[i]); memset(c, 0, sizeof(c)); for(int i = 0; i < k; ++i){ int u, v, w; scanf("%d%d%d", &u, &v, &w); c[u-1][v-1] = w; } b[0] = 1; for(int i = 1; i <= n; ++i) b[i] = b[i-1]*2; for(int i = 0; i < n; ++i) dp[b[i]][i] = a[i]; for(int mask = 0; mask < b[n]; mask++) for(int i = 0; i < n; ++i) if(mask&b[i]){ // 再使用当前状态时,最优解一定已经得出; for(int j = 0; j < n; ++j) if((mask&b[j]) == 0) dp[mask|b[j]][j] = max(dp[mask|b[j]][j], dp[mask][i]+c[i][j]+a[j]); } long long ans = 0; for(int mask = 0; mask < b[n]; mask++) if(len(mask) == m) for(int i = 0; i < n; ++i) ans = max(ans, dp[mask][i]); printf("%I64d\n", ans); }}
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