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。。

/*
   State:Accepted
   Time:2013-03-19 19:26:02
*/
#include<cstring>
#include<string>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<set>
using namespace std;
struct oo{  int x, y; };
int N, K , B ;
const int num[5] = {0, 1, 1, 2, 2};
oo  a[1500];
int f[1010][1010][5] , map[1500] ,p[1500],tot(0);
void init(){
    scanf("%d%d%d",&N, &K, &B);
    for (int i = 1; i <= N; ++i )
       scanf("%d%d",&a[i].y, &a[i].x);
}

int cmp(const oo a, const oo b){
    if (a.x < b.x || a.x == b.x && a.y < b.y ) return true;
    return false;
}

void solve(){
    sort(a + 1, a + 1 + N, cmp);
    memset(map, 0 ,sizeof(map));
    for (int i = 1; i <= N;  ++i){
             if (a[i].x != a[i -1].x) ++tot;
             map[tot]= map[tot] | 1 << a[i].y - 1;
             p[tot] = a[i].x;
      }
      
    for (int i = 0; i <= tot; ++i)
        for (int j = 0; j <= K; ++j)
           for (int opt = 0; opt <= 4; ++opt)
             f[i][j][opt] = 0xfffffff;

    f[0][0][0] = 0;       
    for (int  i = 1; i <= tot; ++i)
         for (int j = 1; j <= K; ++j)
             for (int opt = 1; opt <= 4; ++opt)
               if ((opt & map[i]) >= map[i] || opt == 4){
                     f[i][j][opt] = min(f[i-1][j][opt], f[i-1][j][4]) +  (p[i] - p[i-1]) * num[opt];
                     if (opt == 4){
                           f[i][j][4] = min(f[i][j][4],min(f[i-1][j-1][1], f[i-1][j-1][2]) 
                                                       + p[i] - p[i-1] + 1);
                     }
                     
                     for (int nopt = 0; nopt <= 4; ++nopt){
                       if (opt == 4){
                             if (j > 1) f[i][j][opt] = min(f[i-1][j-2][nopt] + 2,f[i][j][opt]) ;
                        }  else   f[i][j][opt] = min( f[i][j][opt],f[i-1][j-1][nopt] + num[opt]);

                     }
               }
   /* int ans = 0xfffffff;
     for (int  i = 1; i <= tot; ++i)
         for (int j = 1; j <= K; ++j)
             for (int opt = 1; opt <= 4; ++opt)
                if (f[i][j][opt] < 0xfffffff) printf("f[%d][%d][%d] = %d \n",i , j , opt, f[i][j][opt]);*/
    for (int i = 0; i <= 4; ++i)
       ans = min(f[tot][K][i] , ans);
    printf("%d\n",ans);
}

int main(){
    freopen("poj2430.in","r",stdin);
    freopen("poj2430.out","w",stdout);
    init();
    solve();
    fclose(stdin); fclose(stdout);
}