Hdu1081:to the max (maximum sub-matrix, linear DP)

Title: http://acm.hdu.edu.cn/showproblem.php?pid=1081

Oneself really enough rubbish, obviously did one dimension of this problem, but encountered two-dimensional this topic, unexpectedly will not, I also served (PS: pig AH).

Finally, I looked at the puzzle.

The code is as follows:

#include <iostream>#include<stdio.h>#include<string.h>#include<algorithm>#defineINF 0x3f3f3f3fusing namespacestd;intn,w[ the][ the],t;intMain () { while(SCANF ("%d", &n)! =EOF) {memset (W,0,sizeof(w));  for(intI=1; i<=n;i++)        {             for(intj=1; j<=n;j++) {scanf ("%d",&t); W[I][J]+=w[i][j-1]+T; }        }        intmaxx=-inf;  for(intI=1; i<=n;i++)        {             for(intj=i;j<=n;j++)            {                intsum=0;  for(intk=1; k<=n;k++)                {                    if(sum<0) sum=0; Sum+=w[k][j]-w[k][i-1]; if(sum>Maxx) Maxx=sum; }}} printf ("%d\n", Maxx); }    return 0;}

The main puzzle of the Great God:

The problem is that the largest field of one dimension is extended to two dimensions, which is done in the process of finding the largest field in one dimension:

intMax_sum (intN) {    intI, j, sum =0, max =-10000;  for(i =1; I <= N; i++)    {        if(Sum <0) Sum=0; Sum+=A[i]; if(Sum >max) Max=sum; }    returnsum;}

Extended to two-dimensional time is the same method, but need to be two-dimensional compression into one dimension, so we have to do the data processing, so that map[i][j] from the expression of the first line of the J element to represent the first J elements and, so map[k][j]-map[k][i] It is possible to represent the elements of the K row from the I->j column and. As long as the Villivido two-layer loop enumeration I and J is ok.

#include <iostream>#defineMAX 101using namespacestd;intMap[max][max];intMain () {intN, I, J, temp, K;  while(SCANF ("%d", &n)! =EOF) {memset (map,0,sizeof(map));  for(i =1; I <= N; i++)             for(j =1; J <= N; J + +) {scanf ("%d", &temp); MAP[I][J]+ = Map[i][j-1] + temp;//This represents the former J list of line I and            }        intmax =-100000;  for(i =1; I <= N; i++)             for(j = i; J <= N; j + +)            {                intsum =0;  for(k =1; K <= N; k++)                {                    if(Sum <0) Sum=0; Sum+ = Map[k][j]-map[k][i-1];//this represents the former K-line, i->j list and                    if(Sum >max) Max=sum; }} printf ("%d\n", Max); }return 0;}

Hdu1081:to the max (maximum sub-matrix, linear DP)