動態規劃---狀壓dp2
阿新 • • 發佈:2018-08-08
mes getc 今天 png 附加 cor ons image namespace
今天模擬,狀壓dp又沒寫出來。。。還是不會啊,所以今天搞一下這個狀壓dp。這裏有一道狀壓dp的板子題:
Corn Fields
就是一道很簡單的狀壓裸題,但是要每次用一個二進制數表示一行的狀態。
附加一個關於位運算的總結:
上題幹:
題目描述 Farmer John has purchased a lush new rectangular pasture composed of M by N (1 ≤ M ≤ 12; 1 ≤ N ≤ 12) square parcels. He wants to grow some yummy corn for the cows on a number of squares. Regrettably, some of the squares are infertile and can‘t be planted. Canny FJ knows that the cows dislike eating close to each other, so when choosing which squares to plant, he avoids choosing squares that are adjacent; no two chosen squares share an edge. He has not yet made the final choice as to which squares to plant. Being a very open-minded man, Farmer John wants to consider all possible options forhow to choose the squares for planting. He is so open-minded that he considers choosing no squares as a valid option! Please help Farmer John determine the number of ways he can choose the squares to plant. 農場主John新買了一塊長方形的新牧場,這塊牧場被劃分成M行N列(1 ≤ M ≤ 12; 1 ≤ N ≤ 12),每一格都是一塊正方形的土地。John打算在牧場上的某幾格裏種上美味的草,供他的奶牛們享用。 遺憾的是,有些土地相當貧瘠,不能用來種草。並且,奶牛們喜歡獨占一塊草地的感覺,於是John不會選擇兩塊相鄰的土地,也就是說,沒有哪兩塊草地有公共邊。 John想知道,如果不考慮草地的總塊數,那麽,一共有多少種種植方案可供他選擇?(當然,把新牧場完全荒廢也是一種方案) 輸入輸出格式 輸入格式: 第一行:兩個整數M和N,用空格隔開。 第2到第M+1行:每行包含N個用空格隔開的整數,描述了每塊土地的狀態。第i+1行描述了第i行的土地,所有整數均為0或1,是1的話,表示這塊土地足夠肥沃,0則表示這塊土地不適合種草。 輸出格式: 一個整數,即牧場分配總方案數除以100,000,000的余數。 輸入輸出樣例 輸入樣例#1: 復制 2 3 1 1 1 0 1 0 輸出樣例#1: 復制 9
題目不用多解釋,直接上代碼,寫註釋了,很好懂。
#include<cstdio> #include<cstring> using namespace std; #define duke(i,a,n) for(int i = a;i <= n;i++) const int mod = 1e9; template <class T> void read(T &x) { char c; bool op = 0; while(c = getchar(),c > ‘9‘ || c < ‘0‘) if(c == ‘-‘) op = 1; x = c - ‘0‘; while(c = getchar(),c <= ‘9‘&& c >= ‘0‘) x = x * 10 + c - ‘0‘; if(op == 1) x = -x; } int F[15],f[15][4100],m,n,field[15][15]; int state[4110]; int main() { read(m);read(n); duke(i,1,m) duke(j,1,n) read(field[i][j]); duke(i,1,m) { duke(j,1,n) { F[i] = (F[i] << 1) + field[i][j];//F存i行草地的情況 } } f[0][0] = 1; int maxn = 1 << n; duke(i,0,maxn - 1) { state[i] = (((i << 1) & i) == 0) && (((i >> 1) & i) == 0); //每種狀態是否可行 } duke(i,1,m) duke(j,0,maxn - 1) if(state[j] && ((j & F[i]) == j)) //j是否能選 duke(k,0,maxn - 1) //上一排能選什麽 if((j & k) == 0) f[i][j] = (f[i][j] + f[i - 1][k]) % mod; int ans = 0; duke(i,0,maxn - 1) { ans += f[m][i]; ans %= mod; } printf("%d\n",ans); return 0; }
動態規劃---狀壓dp2