poj 1414 DP

来源：互联网发布：网络中的涵化理论编辑：程序博客网时间：2024/06/07 05:35
网上搜的分类是DP，但是遍历就行了。。。。只不过有点麻烦。。。。。
AC代码如下：
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;#define MAX 0x3f3f3f3fstruct Point{int x, y;};struct Group{int id;int size;Point p[55];int cnt;};int num[11][11], N, C, groupnum;Group group[55];bool mark[11][11];int hashs[11][11];int moves[][2] = { { 0, 1 }, { 0, -1 }, { -1, -1 }, { -1, 0 }, { 1, 0 }, { 1, 1 } };void DFS( int x, int y, int groupid, queue<Point> &q ){if( x > N || x < 1 || y > N || y < 1 || y > x || mark[x][y] ){return;}if( num[x][y] != group[groupid].id ){return;}Point pp;pp.x = x;pp.y = y;q.push( pp );mark[pp.x][pp.y] = true;for( int i = 0; i < 6; i++ ){DFS( x + moves[i][0], y + moves[i][1], groupid, q );}}int main(){while( cin >> N >> C && !( N == 0 && C == 0 ) ){groupnum = 0;//读数据for( int i = 1; i <= N; i++ ){for( int j = 1; j <= i; j++ ){scanf( "%d", &num[i][j] );}}//分组memset( mark, false, sizeof( mark ) );for( int i = 1; i <= N; i++ ){for( int j = 1; j <= i; j++ ){if( num[i][j] != 0 && !mark[i][j] ){group[groupnum].id = num[i][j];int cnt = 0;queue<Point> q;while( !q.empty() ){q.pop();}DFS( i, j, groupnum, q );//深搜搜出相连的为一组group[groupnum].size = q.size();//找出与该组相连的空位bool tempmark[130];memset( tempmark, false, sizeof( tempmark ) );while( !q.empty() ){Point p = q.front();q.pop();hashs[p.x][p.y] = groupnum;for( int k = 0; k < 6; k++ ){Point temppoint = p;temppoint.x += moves[k][0];temppoint.y += moves[k][1];if( temppoint.x > N || temppoint.x < 1 || temppoint.y > N || temppoint.y < 1 || temppoint.y > temppoint.x ){continue;}if( num[temppoint.x][temppoint.y] != 0 || tempmark[temppoint.x*11+temppoint.y] ){continue;}group[groupnum].p[cnt].x = temppoint.x;group[groupnum].p[cnt++].y = temppoint.y;tempmark[temppoint.x*11+temppoint.y] = true;}}group[groupnum++].cnt = cnt;}}}//遍历int ans = -MAX;for( int i = 1; i <= N; i++ ){for( int j = 1; j <= i; j++ ){if( num[i][j] != 0 ){continue;}int sub = 1, add = 0;//计算能消的for( int k = 0; k < groupnum; k++ ){if( group[k].cnt > 1 ){continue;}else{if( group[k].p[0].x == i && group[k].p[0].y == j ){if( group[k].id == C ){sub += group[k].size;}else{add += group[k].size;}}}}//计算自己要被消的数量int flag = 1;for( int k = 0; k < 6; k++ ){int tempx = i, tempy = j;tempx += moves[k][0];tempy += moves[k][1];if( tempx > N || tempx < 1 || tempy > N || tempy < 1 || tempy > tempx ){continue;}if( num[tempx][tempy] == 0 ){flag = 0;break;}if( num[tempx][tempy] == C && group[hashs[tempx][tempy]].cnt > 1 ){flag = 0;break;}}if( !flag ){sub = 0;}//取最大ans = max( ans, add - sub );}}cout << ans << endl;}return 0;}
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