HDU 2167 Pebbles

#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <utility>
using namespace std;

#define rep(i,j,k) for (int i=j;i<=k;i++)
#define Rrep(i,j,k) for (int i=j;i>=k;i--)

#define Clean(x,y) memset(x,y,sizeof(x))
#define LL long long
#define ULL unsigned long long
#define inf 0x7fffffff
#define mod %100000007

int n;
int uplim;
vector<int> state;

char str[1009];
int a[20][20];
int dp[16][1600];

bool check(int x)
{
if ( x & (x<<1) || x & (x>>1) ) return false;
return true;
}

int cal( int row , int s ) //计算第row行的数，选取状态为s时的得分
{
int ans = 0;
int k = 0;
while( s )
{
k++;
if ( s & 1 )
{
ans+=a[row][k];
}
s>>=1;
}
return ans;
}

bool init() //读入和预处理
{
int x;
string temp;
getline(cin,temp);
stringstream ss(temp);
n = 0;
while( ss>>x )
{
n++;
a[1]
= x;
}
if ( !n ) return false;
rep(i,2,n)
rep(j,1,n) scanf("%d",&a[i][j]);
getchar();
getchar();
uplim = 1<<n;
Clean(dp,0);
state.clear();
rep(i,0,uplim-1)  //将所有可行状态筛选出来
if ( check(i) ) state.push_back(i);
rep(i,0,state.size()-1) dp[1][i] = cal( 1 , state[i] ); //初始化第一行的值
return true;
}

bool can(int x,int y)
{
if ( x & y ) return false;
if ( x & (y>>1) ) return false;
if ( x & (y<<1) ) return false;
return true;
}

int main()
{
/*
int sum =0;
uplim = 1<<15;
rep(i,0,uplim-1)
if ( check(i) ) sum++;
cout<<sum<<endl;  //计算最大规模时的可行状态数
*/

while( init() )
{
rep(i,2,n) //枚举行数
rep(j,0,state.size()-1) // 枚举第i行的状态
{
int temp = cal(i,state[j]); //get得分

//printf("%d : %d %d\n",i,state[j],temp);
rep(k,0,state.size()-1)  //枚举上一行的状态
if ( can(state[j] , state[k]) )  //如果这两个状态可以相邻，就更新答案
dp[i][j] = max( dp[i][j] , dp[i-1][k] + temp );
}
int ans = 0;
rep(i,0,state.size()-1) ans = max(ans,dp
[i]); //找到最大值
cout<<ans<<endl;
}

return 0;
}