TCO 2014 Round 1C 概率DP

TCO round 1C的 250 和500 的题目都太脑残了，不说了。 TCO round 1C 950 一个棋子，每次等概率的向左向右移动，然后走n步之后，期望cover的区域大小？求cover，肯定就是dp[l][r]
, 走了n步之后，左边cover了l，右边cover了r。 一开始DP没有搞清楚，这个要画一下图就更清楚了。 转移方程就是概率的传递方向.

double dp[505][505][2];  // l,r,n steps unsed;

class RedPaint

{

public:

double expectedCells(int N)

{

memset(dp,0,sizeof(dp));

// the probability of the configuration.

dp[0][0][0] = 1.0;

dp[0][1][1] = 0.5;

dp[1][0][1] = 0.5;

for(int n = 2; n<N+1; n++)

    {

for(int i=0; i<N+1; i++) for(int j=0; j<N+1; j++) dp[i][j][n&1] = 0;

for(int l= 0; l<N+1; l++)

{

for(int r = 0; r<N+1; r++)

    {

if(l== r && r == 0 ) continue;

if( l == 0) dp[l][r][n&1] = (dp[0][r-1][(n-1)&1] + dp[1][r-1][(n-1)&1]) * 0.5;

else if(r == 0) dp[l][r][n&1] = (dp[l-1][r][(n-1)&1] + dp[l-1][r+1][(n-1)&1])  * 0.5;

else

dp[l][r][n&1] = (dp[max(0,l-1)][r+1][(n-1)&1] + dp[l+1][max(r-1,0)][(n-1)&1]) * 0.5;

}

}

}

double ret = 0.0f;

double pro = 0.0f;

for(int l=0; l<N+1; l++)

    {

for(int r = 0; r<N+1; r++)

{

pro += dp[l][r][N&1];

cout<<l<<" "<<r<<" "<<N<<" "<<dp[l][r][N&1]<<endl;

ret += (l+r+1) * dp[l][r][N&1];

}

}

cout<<"All Pro "<<pro<<endl;

return ret;

}

};

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