poj2430
题目意思：一个2*b的矩形上有n头牛，现在要用k个矩形覆盖牛。问最少覆盖的面积
思路：状态压缩dp
因为b太大，所以得先离散化
f[i][j][k]表示前i列（有牛在的列），用了k个矩形，状态为K所用的最小面积
情况有：1.单独放1个矩形
2.单独放2个矩形
3.由前面放的延伸过来（1个或2个）
情况多了点，写起来挺麻烦的。。
1 /* 2 State:Accepted 3 Time:2013-03-19 19:26:02 4 */ 5 #include<cstring> 6 #include<string> 7 #include<cstdlib> 8 #include<cstdio> 9 #include<cmath> 10 #include<algorithm> 11 #include<set> 12 using namespace std; 13 struct oo{ int x, y; }; 14 int N, K , B ; 15 const int num[5] = {0, 1, 1, 2, 2}; 16 oo a[1500]; 17 int f[1010][1010][5] , map[1500] ,p[1500],tot(0); 18 void init(){ 19 scanf("%d%d%d",&N, &K, &B); 20 for (int i = 1; i <= N; ++i ) 21 scanf("%d%d",&a[i].y, &a[i].x); 22 } 23 24 int cmp(const oo a, const oo b){ 25 if (a.x < b.x || a.x == b.x && a.y < b.y ) return true; 26 return false; 27 } 28 29 void solve(){ 30 sort(a + 1, a + 1 + N, cmp); 31 memset(map, 0 ,sizeof(map)); 32 for (int i = 1; i <= N; ++i){ 33 if (a[i].x != a[i -1].x) ++tot; 34 map[tot]= map[tot] | 1 << a[i].y - 1; 35 p[tot] = a[i].x; 36 } 37 38 for (int i = 0; i <= tot; ++i) 39 for (int j = 0; j <= K; ++j) 40 for (int opt = 0; opt <= 4; ++opt) 41 f[i][j][opt] = 0xfffffff; 42 43 f[0][0][0] = 0; 44 for (int i = 1; i <= tot; ++i) 45 for (int j = 1; j <= K; ++j) 46 for (int opt = 1; opt <= 4; ++opt) 47 if ((opt & map[i]) >= map[i] || opt == 4){ 48 f[i][j][opt] = min(f[i-1][j][opt], f[i-1][j][4]) + (p[i] - p[i-1]) * num[opt]; 49 if (opt == 4){ 50 f[i][j][4] = min(f[i][j][4],min(f[i-1][j-1][1], f[i-1][j-1][2]) 51 + p[i] - p[i-1] + 1); 52 } 53 54 for (int nopt = 0; nopt <= 4; ++nopt){ 55 if (opt == 4){ 56 if (j > 1) f[i][j][opt] = min(f[i-1][j-2][nopt] + 2,f[i][j][opt]) ; 57 } else f[i][j][opt] = min( f[i][j][opt],f[i-1][j-1][nopt] + num[opt]); 58 59 } 60 } 61 /* int ans = 0xfffffff; 62 for (int i = 1; i <= tot; ++i) 63 for (int j = 1; j <= K; ++j) 64 for (int opt = 1; opt <= 4; ++opt) 65 if (f[i][j][opt] < 0xfffffff) printf("f[%d][%d][%d] = %d \n",i , j , opt, f[i][j][opt]);*/ 66 for (int i = 0; i <= 4; ++i) 67 ans = min(f[tot][K][i] , ans); 68 printf("%d\n",ans); 69 } 70 71 int main(){ 72 freopen("poj2430.in","r",stdin); 73 freopen("poj2430.out","w",stdout); 74 init(); 75 solve(); 76 fclose(stdin); fclose(stdout); 77 }
