题目：http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1522
很好的思想。考虑从小到大一对一对填数，这样也能对它的大小限制做一些操作了。
因为从小到大，所以只能全填在左边、全填在右边、两边各填一个。记录左边填到了哪个位置，就可知右边填到了哪个位置。转移之前判断一下这样填是否合法即可。
新的不合法的状态只会和现在填的两个位置有关。
注意输入格式！！符号前后有空格！！！
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #define ll long long using namespace std; const int N=40,M=105; int n,m,x[M],y[M],sgn[M],tx,ty; ll dp[N][N<<1],ans; char ch[20]; bool check(int p0,int p1,int r,int fx) {//printf("p0=%d p1=%d r=%d fx=%d\n",p0,p1,r,fx);for(int i=1;i<=m;i++)if(sgn[i]==0&&( ( (x[i]==p0||x[i]==p1)&&y[i]!=p0&&y[i]!=p1)||( (y[i]==p0||y[i]==p1)&&x[i]!=p0&&x[i]!=p1) ) )return 0;//以前都合法，不合法仅出现在p0、p1位置上if(!fx){for(int i=1;i<=m;i++){//printf("sgn[%d]=%d\n",i,sgn[i]);if(sgn[i]==1||sgn[i]==2){tx=x[i];ty=y[i];if(sgn[i]==2)swap(tx,ty);//printf("tx=%d ty=%d p0=%d p1=%d r=%d\n",tx,ty,p0,p1,r);if((ty==p0||ty==p1)&&tx>=p0&&tx<r) return 0;}if(sgn[i]==3||sgn[i]==4){tx=x[i];ty=y[i];if(sgn[i]==4)swap(tx,ty);if((ty==p0||ty==p1)&&tx>p1&&tx<r) return 0;}}}if(fx==1){for(int i=1;i<=m;i++){if(sgn[i]==1||sgn[i]==2){tx=x[i];ty=y[i];if(sgn[i]==2)swap(tx,ty);if((ty==p0||ty==p1)&&tx<=p1&&tx>r) return 0;}if(sgn[i]==3||sgn[i]==4){tx=x[i];ty=y[i];if(sgn[i]==4)swap(tx,ty);if((ty==p0||ty==p1)&&tx<p0&&tx>r) return 0;}}}if(fx==2){for(int i=1;i<=m;i++){if(sgn[i]==1||sgn[i]==2){tx=x[i];ty=y[i];if(sgn[i]==2)swap(tx,ty);if((ty==p0||ty==p1)&&tx>=p0&&tx<=p1) return 0;}if(sgn[i]==3||sgn[i]==4){tx=x[i];ty=y[i];if(sgn[i]==4)swap(tx,ty);if((ty==p0||ty==p1)&&tx>p0&&tx<p1) return 0;}}}return 1; } int main() {scanf("%d%d",&n,&m);ch[0]=getchar();while(ch[0]!='\n')ch[0]=getchar();for(int i=1,len=0,j;i<=m;i++,len=0){ch[++len]=getchar();while(ch[len]!='\n')ch[++len]=getchar();for(j=1;j<len;j++){if(ch[j]>='0'&&ch[j]<='9')x[i]=(x[i]<<3)+(x[i]<<1)+ch[j]-'0';else break;}while(ch[j]==' ')j++;//printf("j=%d chj=(%c)\n",j,ch[j]);if(ch[j]=='=') sgn[i]=0,j++;else if(ch[j]=='<'&&ch[j+1]=='=')sgn[i]=3,j+=2;else if(ch[j]=='>'&&ch[j+1]=='=')sgn[i]=4,j+=2;else if(ch[j]=='<')sgn[i]=1,j++;else if(ch[j]=='>')sgn[i]=2,j++;while(ch[j]==' ')j++;//printf("j=%d chj=(%c)\n",j,ch[j]);for(;j<len;j++)y[i]=(y[i]<<3)+(y[i]<<1)+ch[j]-'0';//printf("x=%d sgn=%d y=%d\n",x[i],sgn[i],y[i]); }dp[0][0]=1;for(int i=0,lm;i<n;i++){lm=(i<<1);for(int j=0,r;j<=lm;j++){//if(dp[i][j])printf("dp[%d][%d]=%lld\n",i,j,dp[i][j]);r=(n<<1)-(lm-j)+1;if(check(j+1,j+2,r,0))dp[i+1][j+2]+=dp[i][j];//printf("dp[%d][%d]=%lld(%d,%d)\n",i+1,j+2,dp[i+1][j+2],i,j);if(check(r-2,r-1,j,1))dp[i+1][j]+=dp[i][j];//printf("dp[%d][%d]=%lld(%d,%d)\n",i+1,j,dp[i+1][j],i,j);if(check(j+1,r-1,0,2))dp[i+1][j+1]+=dp[i][j];//printf("dp[%d][%d]=%lld(%d,%d)\n",i+1,j+1,dp[i+1][j+1],i,j); }}int lm=(n<<1);for(int j=0;j<=lm;j++) ans+=dp[n][j];printf("%lld\n",ans/3);return 0; }
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