UVA 10593 Kites DP

The season of Ying Kites is a well ahead. So what? Let us make a inventory for kites. We are given
A square shaped sheet of paper. But many parts of this is already porous. Your Challenge here's to
The count of the total number of ways to cut a kite of any size from this sheet. By the kite itself
Can ' t be porous:-) And ... it must be either square shaped or diamond shaped.
X
x xxx xxx xxx
xxx xxxxx XXX x.x x
x xxx xxx xxx
X
In the above gure rst three is valid kites but not next.
Input
Input contains an integer n (n≤500), which is the size of the sheet. Then follows n lines each of the which
has n characters (' X ' or '. '). Here the dotted parts resemble the porous parts of the sheet. Input is
Terminated by end of Le.
Output
Output is very simple. Only print a integer according to the problem statement for each test case in
A new line.
Sample Input
4
. xx.
Xxxx
. xx.
. x..
3
Xxx
Xxx
Xxx
Sample Output
4
6

Test Instructions : give you a n*n figure that lets you figure out the X-shaped side of the graph and grow to a diamond or square that equals 2.

A: DP, this blog is better than the figure

http://blog.csdn.net/u012596172/article/details/41171815

//Meek#include <bits/stdc++.h>#include<iostream>#include<cstdio>#include<cmath>#include<string>#include<cstring>#include<algorithm>#include<map>#include<queue>using namespacestd; typedefLong Longll;#defineMem (a) memset (A,0,sizeof (a))#definePB Push_back#defineFi first#defineSe Second#defineMP Make_pairConst intn=550;Constll INF = 1ll<< A;Const intINF =1000000007;Const intMOD =2000000011;CharMp[n][n];intDp[n][n],n;intsolve () {intAns =0;    MEM (DP);  for(intI=1; i<=n;i++) {         for(intj=1; j<=n;j++) {            if(Mp[i][j] = ='x') {                inttmp = MIN (dp[i-1][j],dp[i][j-1]); DP[I][J]= tmp + (mp[i-tmp][j-tmp] = ='x'); if(Dp[i][j] >1) ans +=dp[i][j]-1;    }}} mem (DP);  for(intI=1; i<=n;i++) {         for(intj=1; j<=n;j++) {            if(Mp[i][j] = ='x') {                inttmp = MIN (dp[i-1][j-1],dp[i-1][j+1]); if(TMP = =0|| mp[i-1][J]! ='x') Dp[i][j] =1; Else if(mp[i-2*tmp][j]=='x'&& mp[i-tmp*2+1][J] = ='x') Dp[i][j] = tmp +1; ElseDP[I][J] =tmp; if(Dp[i][j] >1) ans +=dp[i][j]-1; }        }    }    returnans;}intMain () { while(SCANF ("%d", &n)! =EOF) {         for(intI=1; i<=n;i++) {GetChar ();  for(intj=1; j<=n;j++) scanf ("%c",&Mp[i][j]); } printf ("%d\n", Solve ()); }    return 0;}

Code

UVA 10593 Kites DP