SPOJ 274 Johnny and the Watermelon Plantation（TLE）

O(n^3)的时间复杂度，改了半天交了二三十遍，TLE到死，实在没办法了……

跪求指点！！！

  1 #include <cstdio>

  2 #include <cstdlib>

  3 #include <cstring>

  4 #include <algorithm>

  5 

  6 using namespace std;

  7 

  8 const int MAXN = 1010;

  9 const int INF = 1 << 30;

 10 

 11 int mat[MAXN][MAXN];

 12 int SubSum[MAXN][MAXN];

 13 int tempS[MAXN];

 14 int N, K;

 15 int maxI, maxJ;

 16 int hang[MAXN], cntH;

 17 int lie[MAXN], cntL;

 18 

 19 //预处理左上角[1,1]右下角[i,j]的所有矩阵和

 20 void init()

 21 {

 22     memset( SubSum, 0, sizeof(SubSum) );

 23     for ( int i = 1; i <= maxI; ++i )

 24     {

 25         int tmps = 0;

 26         for ( int j = 1; j <= maxJ; ++j )

 27         {

 28             tmps += mat[i][j];

 29             SubSum[i][j] = SubSum[i - 1][j] + tmps;

 30         }

 31     }

 32     return;

 33 }

 34 

 35 void solved()

 36 {

 37     int ans = INF;

 38     for ( int i = 0; i < cntH; ++i ) //枚举矩形上边界

 39     for ( int j = i; j < cntH; ++j )  //枚举矩形下边界

 40     {

 41         int stL = hang[i], edL = hang[j];

 42         tempS[0] = 0;

 43         int st = 1;

 44         //扫描法，参考刘汝佳训练指南p.48 LA 2678

 45         for ( int m = 1; m <= cntL; ++m )

 46         {

 47             int k = lie[m - 1];

 48             tempS[m] = SubSum[edL][k] - SubSum[edL][k-1] - SubSum[stL-1][k] + SubSum[stL-1][k-1];

 49             tempS[m] += tempS[m - 1];

 50             if ( tempS[st - 1] > tempS[m] - K ) continue;

 51             while ( tempS[st] <= tempS[m] - K ) ++st;

 52             ans = min( ans, ( lie[m - 1] - lie[st - 1] ) * ( edL - stL ) );

 53         }

 54     }

 55     printf( "%d\n", ans );

 56     return;

 57 }

 58 

 59 int main()

 60 {

 61     //freopen( "in.txt", "r", stdin );

 62     int T;

 63     scanf( "%d", &T );

 64     while ( T-- )

 65     {

 66         memset( mat, 0, sizeof(mat) );

 67         scanf( "%d%d", &N, &K );

 68         maxI = 0;

 69         maxJ = 0;

 70         cntL = 0, cntH = 0;

 71         bool flag = false;

 72         for ( int i = 0; i < N; ++i )

 73         {

 74             int x, y, f;

 75             scanf("%d%d%d", &x, &y, &f );

 76             hang[cntH++] = x;  //行列离散化

 77             lie[cntL++] = y;

 78             maxI = max( maxI, x );

 79             maxJ = max( maxJ, y );

 80             mat[x][y] = f;

 81             if ( f >= K )  //特殊情况，只覆盖一个点

 82             {

 83                 flag = true;

 84             }

 85         }

 86 

 87         if ( N == 1 || flag )

 88         {

 89             puts("0");

 90             continue;

 91         }

 92 

 93         sort( hang, hang + cntH );

 94         sort( lie, lie + cntL );

 95         cntH = unique( hang, hang + cntH ) - hang;

 96         cntL = unique( lie, lie + cntL ) - lie;

 97 

 98         init();

 99         solved();

100     }

101     return 0;

102 }