对task和machine的yi由小到大进行排序，然后对machine来跟task配对。当machine[].yi >= task[].yi时，就更新线段树，在1-1440上做线段树,线段树存的是task[].xi，同时用用优先队列保存task[].yi；当machine[].yi < task[].yi时，就查找 1到machine[].xi最大的值。如果存在最大值的话，把优先队列里的task[].yi取出来。。这样一个machine就匹配到了一个最优的任务。还是看代码好好意会吧，细节挺多的
1 #include<iostream> 2 #include<cstring> 3 #include<algorithm> 4 #include<cstdio> 5 #include<string> 6 #include<queue> 7 #include<cmath> 8 #include<map> 9 10 using namespace std; 11 12 #define mnx 100050 13 #define ll long long 14 #define inf 0x3f3f3f3f 15 #define lson l, m, rt << 1 16 #define rson m+1, r, rt << 1 | 1 17 18 priority_queue<int> q[1500]; 19 int cnt[1500], sum[8000]; 20 struct s{ 21 int xi, yi; 22 bool operator < ( const s & b ) const{ 23 return yi < b.yi; 24 } 25 }a[mnx], b[mnx]; 26 void pushup( int rt ){ 27 sum[rt] = max( sum[rt<<1], sum[rt<<1|1] ); 28 } 29 void build( int l, int r, int rt ){ 30 if( l == r ){ 31 sum[rt] = -1; 32 return ; 33 } 34 int m = ( l + r ) >> 1; 35 build( lson ), build( rson ); 36 pushup( rt ); 37 } 38 int find( int L, int R, int l, int r, int rt ){ 39 if( L <= l && R >= r ){ 40 return sum[rt]; 41 } 42 int m = ( l + r ) >> 1; 43 int ret = -1; 44 if( L <= m ) ret = max( ret, find( L, R, lson ) ); 45 if( R > m ) ret = max( ret, find( L, R, rson ) ); 46 return ret; 47 48 } 49 void update( int u, int v, int l, int r, int rt ){ 50 if( l == r ){ 51 sum[rt] = v; 52 return ; 53 } 54 int m = ( l + r ) >> 1; 55 if( u <= m ) update( u, v, lson ); 56 else update( u, v, rson ); 57 pushup( rt ); 58 } 59 int main(){ 60 int n, m; 61 const int nn = 1450; 62 while( scanf( "%d%d", &n, &m ) != EOF ){ 63 for( int i = 0; i < nn; i++ ){ 64 while( !q[i].empty() ) q[i].pop(); 65 } 66 memset( cnt, 0, sizeof(cnt) ); 67 for( int i = 0; i < n; i++ ){ 68 scanf( "%d%d", &a[i].xi, &a[i].yi ); 69 } 70 for( int i = 0; i < m; i++ ){ 71 scanf( "%d%d", &b[i].xi, &b[i].yi ); 72 } 73 sort( a, a+n ); 74 sort( b, b+m ); 75 build( 1, nn, 1 ); 76 int j = 0, ans1 = 0; ll ans2 = 0; 77 for( int i = 0; i < n; i++ ){ 78 while( j < m && a[i].yi >= b[j].yi ){ 79 if( !cnt[b[j].xi] ){ 80 update( b[j].xi, b[j].xi, 1, nn, 1 ); 81 } 82 cnt[b[j].xi]++; 83 q[b[j].xi].push( b[j].yi ); 84 j++; 85 } 86 int t = find( 1, a[i].xi, 1, nn, 1 ); 87 if( t == - 1 ) continue; 88 cnt[t]--; 89 if( cnt[t] == 0 ) update( t, -1, 1, nn, 1 ); 90 int tt = q[t].top(); q[t].pop(); 91 ans1++; 92 ans2 += t * 500 + tt * 2; 93 } 94 printf( "%d %I64d ", ans1, ans2 ); 95 } 96 return 0; 97 }