Title Link: http://poj.org/problem?id=2482
It's a little sad to finish reading the question (?????? Test instructions: There are n stars (the coordinates of the star I are XI, Yi, Brightness is ci), give you a w*h rectangle, let you find the rectangle can cover the brightness of the stars and the maximum of how much
Idea: The rectangular size is fixed, so you can think in a different direction, the rectangle as a point (coordinates are the center of the rectangle), each star's influence interval range is w*h rectangle (the original coordinates of the star is the center of the rectangle), the brightness of the interval increased C. The problem becomes the one that asks which point is the most luminance. The coordinate range is too large and the direct brute force enumeration is not possible, so it needs to be discretized beforehand. Create a line segment tree vertically, and then enumerate the horizontal axis, updating the vertical coordinate interval that needs to be updated each time on the segment tree, and then querying.
#include <iostream>#include<cstring>#include<algorithm>#defineMAXN 10010#defineINF 1000000000#defineLL (x) x<<1#defineRR (x) x<<1|1using namespaceStd;typedefLong LongLL;//variable definestructline{intx, y1, y2, Light;};structtree{intL, R; intadd; LL Ma;}; Tree NODE[MAXN*8]; LL W, H, STARX[MAXN], STARY[MAXN], XX[MAXN*2], yy[maxn*2];intLIGHT[MAXN], n;line li[maxn*4];//function DefinevoidPush_down (intx);voidPUSH_UP (intx);voidBuild_tree (intLeftintRightintx); LL Query (intLeftintRightintx);voidUpdate_add (intLeftintRightintx, LL val);BOOLLine_compare (line L1, line L2);intMainvoid){ while(SCANF ("%d%lld%lld", &n, &w, &h)! =EOF) { for(inti =0; I < n; ++i) {scanf ("%lld%lld%d", &starx[i], &stary[i], &Light[i]); Starx[i]*=2; Stary[i]*=2; } for(inti =0; I < n; ++i) {xx[i*2] = Starx[i]-W; Xx[i*2+1] = Starx[i] +W; Yy[i*2] = Stary[i]-H; Yy[i*2+1] = Stary[i] + H-1; } sort (xx, xx+2*N); Sort (yy, yy+2*N); for(inti =0; I < n; ++i) {intx, y1, y2; X= (int) (Lower_bound (xx, XX +2*n, Starx[i]-W)-xx); Y1= (int) (Lower_bound (yy, yy +2*n, Stary[i]-H)-yy); Y2= (int) (Lower_bound (yy, yy +2*n, Stary[i] + H-1) -yy); Li[i*2].x =x; Li[i*2].y1 =Y1; Li[i*2].y2 =Y2; Li[i*2].light =Light[i]; X= (int) (Lower_bound (xx, XX +2*n, Starx[i] + W)-xx); Li[i*2+1].x =x; Li[i*2+1].y1 =Y1; Li[i*2+1].y2 =Y2; Li[i*2+1].light =-1*Light[i]; } build_tree (1,2*n,1); LL ans=0; Sort (Li, Li+2*N, Line_compare); for(inti =0; I <2*n; ++i) {Update_add (li[i].y1+1, Li[i].y2 +1,1, Li[i].light); Ans= max (ans, query (li[i].y1 +1, Li[i].y2 +1,1)); } printf ("%lld\n", ans); } return 0;}voidBuild_tree (intLeftintRightintx) {NODE[X].L=Left ; NODE[X].R=Right ; Node[x].add= Node[x].ma =0; if(left = =Right )return; intLX =LL (x); intRx =RR (x); intMid = left + (right-left)/2; Build_tree (left, mid, LX); Build_tree (Mid+1, right, RX); PUSH_UP (x);}voidPUSH_UP (intx) { if(Node[x].l >=NODE[X].R)return; intLX =LL (x); intRx =RR (x); Node[x].ma=Max (node[lx].ma, node[rx].ma);}voidPush_down (intx) { if(Node[x].l >=NODE[X].R)return; intLX =LL (x); intRx =RR (x); if(Node[x].add! =0) {Node[lx].add+=Node[x].add; Node[rx].add+=Node[x].add; Node[lx].ma+=Node[x].add; Node[rx].ma+=Node[x].add; }}voidUpdate_add (intLeftintRightintx, LL val) { if(NODE[X].L = = Left && NODE[X].R = =Right ) {Node[x].add+=Val; Node[x].ma+=Val; return; } push_down (x); Node[x].add=0; intLX =LL (x); intRx =RR (x); intMID = Node[x].l + (NODE[X].R-NODE[X].L)/2; if(Right <=mid) Update_add (left, right, LX, Val); Else if(Left >mid) Update_add (left, right, Rx, Val); Else{Update_add (left, Mid, LX, Val); Update_add (Mid+1, right, Rx, Val); } push_up (x);} LL Query (intLeftintRightintx) { if(NODE[X].L = = Left && NODE[X].R = =Right ) { returnnode[x].ma; } push_down (x); Node[x].add=0; intMID = Node[x].l + (NODE[X].R-NODE[X].L)/2; intLX =LL (x); intRx =RR (x); LL result; if(Right <=mid) Result=query (left, right, LX); Else if(Left >mid) Result=query (left, right, RX); Elseresult= max (query (left, Mid, LX), query (mid +1, right, Rx)); PUSH_UP (x); returnresult;}BOOLLine_compare (line L1, line L2) {if(l1.x! =l2.x)returnl1.x <l2.x; returnL1.light <l2.light;}
POJ 2482 Stars in Your Window (segment tree: Interval update)