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ZOJ 3802 Easy 2048 Again 状压DP

直接从前往后DP,因为一共只有500个数,所以累加起来的话单个数不会超过4096,并且因为是Flappy 2048的规则,所以只有之后数列末尾一串递减的是有效的,因此可以状压。

1700ms = =,据说用滚动数组优化一下会好很多

#include <cstdio>#include <cstring>#include <iostream>#include <map>#include <set>#include <vector>#include <string>#include <queue>#include <deque>#include <bitset>#include <list>#include <cstdlib>#include <climits>#include <cmath>#include <ctime>#include <algorithm>#include <stack>#include <sstream>#include <numeric>#include <fstream>#include <functional>using namespace std;#define MP make_pair#define PB push_backtypedef long long LL;typedef unsigned long long ULL;typedef vector<int> VI;typedef pair<int,int> pii;const int INF = INT_MAX / 3;const double eps = 1e-8;const LL LINF = 1e17;const double DINF = 1e60;const int maxn = 505;const int maxs = (4096 + 5);int num[maxn],n;int f[maxn][maxs];inline int lowbit(int x) {    return x & -x;}void solve() {    memset(f,-1,sizeof(f));    f[0][0] = 0;    int ans = 0;    for(int i = 1;i <= n;i++) {        for(int j = 0;j < 4096;j++) if(f[i - 1][j] != -1) {            int nownum = num[i], nowst = j, nowval = nownum;            f[i][j] = max(f[i][j], f[i - 1][j]);            while(lowbit(nowst) == (nownum >> 1)) {                nowval += (nownum << 1);                nowst ^= (nownum >> 1);                nownum <<= 1;            }            if(lowbit(nowst) < (nownum >> 1)) nowst = 0;            nowst |= (nownum >> 1);            f[i][nowst] = max(f[i][nowst],f[i - 1][j] + nowval);        }    }    for(int i = 0;i < 4096;i++) ans = max(ans,f[n][i]);    printf("%d\n",ans);}int main() {    int T; scanf("%d",&T);    while(T--) {       scanf("%d",&n);       for(int i = 1;i <= n;i++) {           scanf("%d",&num[i]);       }       solve();    }    return 0;}

 

ZOJ 3802 Easy 2048 Again 状压DP