没有正确分析路径可能的条数，它是指数增长的，会爆long long。
然后就是正反两次时间分治。
另一个就是max with count，即带计数的最值，即除了记录最值，还要记录最值取得的次数。
1 /************************************************************** 2 Problem: 2244 3 User: idy002 4 Language: C++ 5 Result: Accepted 6 Time:4108 ms 7 Memory:4520 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <algorithm> 12 #define oo 0x3f3f3f3f 13 #define N 50010 14 using namespace std; 15 16 typedef bool (*Cmp)( int a, int b ); 17 18 struct Point { 19 int x, y; 20 Point(){} 21 Point( int x, int y ):x(x),y(y){} 22 }; 23 struct Info { 24 int v; 25 long double c; 26 Info(){} 27 Info( int v, long double c ):v(v),c(c){} 28 void add( Info o ) { 29 if( o.v>v ) 30 *this = o; 31 else if( o.v==v ) 32 c += o.c; 33 } 34 }; 35 36 int n; 37 Point pts[N]; 38 Info dp[2][N]; 39 Info bit[N]; 40 int q[N]; 41 Cmp cmp; 42 int disc[N], dtot; 43 long double ans[N]; 44 45 bool cmpa( int a, int b ) { 46 if( pts[a].x!=pts[b].x ) return pts[a].x<pts[b].x; 47 if( pts[a].y!=pts[b].y ) return pts[a].y<pts[b].y; 48 return a<b; 49 } 50 bool cmpb( int a, int b ) { 51 if( pts[a].x!=pts[b].x ) return pts[a].x>pts[b].x; 52 if( pts[a].y!=pts[b].y ) return pts[a].y>pts[b].y; 53 return a<b; 54 } 55 void modify( int pos, Info o ) { 56 if( cmp==cmpb ) pos = dtot+1-pos; 57 for( int i=pos; i<=dtot; i+=i&-i ) 58 bit[i].add(o); 59 } 60 void clear( int pos ) { 61 if( cmp==cmpb ) pos = dtot+1-pos; 62 for( int i=pos; i<=dtot; i+=i&-i ) 63 bit[i] = Info(-oo,0); 64 } 65 Info query( int pos ) { 66 if( cmp==cmpb ) pos = dtot+1-pos; 67 Info o(-oo,0); 68 for( int i=pos; i; i-=i&-i ) 69 o.add(bit[i]); 70 return o; 71 } 72 void cdq( Info dp[], int lf, int rg ) { 73 if( lf==rg ) { 74 dp[lf].add( Info(1,1) ); 75 return; 76 } 77 int mid=(lf+rg)>>1; 78 cdq( dp, lf, mid ); 79 for( int i=lf; i<=rg; i++ ) 80 q[i] = i; 81 sort( q+lf, q+rg+1, cmp ); 82 for( int t=lf; t<=rg; t++ ) { 83 int i=q[t]; 84 if( i<=mid ) { 85 modify( pts[i].y, dp[i] ); 86 } else { 87 Info o = query(pts[i].y); 88 o.v++; 89 dp[i].add( o ); 90 } 91 } 92 for( int t=lf; t<=rg; t++ ) { 93 int i=q[t]; 94 if( i<=mid ) 95 clear( pts[i].y ); 96 } 97 cdq( dp, mid+1, rg ); 98 } 99 int main() { 100 scanf( "%d", &n ); 101 for( int i=1,x,y; i<=n; i++ ) { 102 scanf( "%d%d", &x, &y ); 103 pts[i] = Point(x,y); 104 disc[++dtot] = y; 105 } 106 reverse( pts+1, pts+1+n ); 107 sort( disc+1, disc+1+dtot ); 108 dtot = unique( disc+1, disc+1+dtot ) - disc - 1; 109 for( int i=1; i<=n; i++ ) 110 pts[i].y = lower_bound( disc+1, disc+1+dtot, pts[i].y ) - disc; 111 112 cmp = cmpa; 113 cdq( dp[0], 1, n ); 114 cmp = cmpb; 115 reverse( pts+1, pts+1+n ); 116 cdq( dp[1], 1, n ); 117 reverse( dp[1]+1, dp[1]+1+n ); 118 119 Info info(-oo,0); 120 for( int i=1; i<=n; i++ ) 121 info.add( dp[0][i] ); 122 printf( "%d ", info.v ); 123 for( int i=1; i<=n; i++ ) { 124 if( dp[0][i].v+dp[1][i].v-1 == info.v ) { 125 ans[i] = dp[0][i].c*dp[1][i].c/info.c; 126 } else 127 ans[i] = 0.0; 128 } 129 reverse( ans+1, ans+1+n ); 130 for( int i=1; i<=n; i++ ) 131 printf( "%.5Lf ", ans[i] ); 132 }