UVA 11992 Fast Matrix Operations (二维线段树)
解法：因为至多20行，所以至多建20棵线段树，每行建一个。具体实现如下，有些复杂，慢慢看吧。
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; #define N 1000010 struct node { int mini,maxi,sum; int addmark,setmark; }tree[21][4*N]; int i; void pushup(int i,int rt) { tree[i][rt].sum = tree[i][2*rt].sum + tree[i][2*rt+1].sum; tree[i][rt].mini = min(tree[i][2*rt].mini,tree[i][2*rt+1].mini); tree[i][rt].maxi = max(tree[i][2*rt].maxi,tree[i][2*rt+1].maxi); } void build(int i,int l,int r,int rt) { tree[i][rt].sum = tree[i][rt].mini = tree[i][rt].maxi = 0; tree[i][rt].setmark = -1; tree[i][rt].addmark = 0; if(l == r) return; int mid = (l+r)/2; build(i,l,mid,2*rt); build(i,mid+1,r,2*rt+1); pushup(i,rt); } void pushdown(int i,int l,int r,int rt) { if(tree[i][rt].setmark == -1 && tree[i][rt].addmark == 0) return; int mid = (l+r)/2; if(tree[i][rt].setmark >= 0) { tree[i][2*rt].sum = tree[i][rt].setmark*(mid-l+1); tree[i][2*rt+1].sum = tree[i][rt].setmark*(r-mid); tree[i][2*rt].mini = tree[i][2*rt+1].mini = tree[i][rt].setmark; // tree[i][2*rt].maxi = tree[i][2*rt+1].maxi = tree[i][rt].setmark; // tree[i][2*rt].addmark = tree[i][2*rt+1].addmark = 0; // tree[i][2*rt].setmark = tree[i][2*rt+1].setmark = tree[i][rt].setmark; tree[i][rt].setmark = -1; } if(tree[i][rt].addmark > 0) { tree[i][2*rt].sum += tree[i][rt].addmark*(mid-l+1); tree[i][2*rt+1].sum += tree[i][rt].addmark*(r-mid); tree[i][2*rt].maxi += tree[i][rt].addmark; // tree[i][2*rt].mini += tree[i][rt].addmark; tree[i][2*rt+1].maxi += tree[i][rt].addmark; tree[i][2*rt+1].mini += tree[i][rt].addmark; // tree[i][2*rt].addmark += tree[i][rt].addmark; tree[i][2*rt+1].addmark += tree[i][rt].addmark; tree[i][rt].addmark = 0; } } void add(int l,int r,int aa,int bb,int val,int rt) { if(aa>r||bb<l) return; if(aa<=l&&bb>=r) { tree[i][rt].addmark += val; //tree[i][rt].setmark = -1; tree[i][rt].sum += (r-l+1)*val; tree[i][rt].maxi += val; tree[i][rt].mini += val; return; } pushdown(i,l,r,rt); int mid = (l+r)/2; if(aa<=mid) add(l,mid,aa,bb,val,2*rt); if(bb>mid) add(mid+1,r,aa,bb,val,2*rt+1); pushup(i,rt); } void setval(int l,int r,int aa,int bb,int val,int rt) { if(aa>r||bb<l) return; if(aa<=l&&bb>=r) { tree[i][rt].setmark = val; tree[i][rt].addmark = 0; tree[i][rt].sum = val*(r-l+1); tree[i][rt].maxi = tree[i][rt].mini = val; return; } pushdown(i,l,r,rt); int mid = (l+r)/2; if(aa<=mid) setval(l,mid,aa,bb,val,2*rt); if(bb>mid) setval(mid+1,r,aa,bb,val,2*rt+1); pushup(i,rt); } struct node_ans { int sum; int mini,maxi; }; node_ans query(int l,int r,int aa,int bb,int rt) { node_ans res,ka1,ka2; if(aa<=l && bb>=r) { res.sum = tree[i][rt].sum; res.maxi = tree[i][rt].maxi; res.mini = tree[i][rt].mini; return res; } pushdown(i,l,r,rt); int mid = (l+r)/2; if(bb<=mid) return query(l,mid,aa,bb,2*rt); else if(aa>mid) return query(mid+1,r,aa,bb,2*rt+1); else { ka1 = query(l,mid,aa,bb,2*rt); ka2 = query(mid+1,r,aa,bb,2*rt+1); res.sum = ka1.sum + ka2.sum; res.maxi = max(ka1.maxi,ka2.maxi); res.mini = min(ka1.mini,ka2.mini); return res; } } int main() { int r,c,m; int k,zuo; int x1,y1,x2,y2,val; int sum,mmax,mmin; while(scanf("%d%d%d",&r,&c,&m)!=EOF) { for(i=1;i<=r;i++) { build(i,1,c,1); } for(k=0;k<m;k++) { scanf("%d",&zuo); if(zuo == 1) { scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&val); for(i=x1;i<=x2;i++) { add(1,c,y1,y2,val,1); } } else if(zuo == 2) { scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&val); for(i=x1;i<=x2;i++) { setval(1,c,y1,y2,val,1); } } else { scanf("%d%d%d%d",&x1,&y1,&x2,&y2); node_ans la; sum = 0; mmax = -Mod; mmin = Mod; for(i=x1;i<=x2;i++) { la = query(1,c,y1,y2,1); sum += la.sum; mmax = max(mmax,la.maxi); mmin = min(mmin,la.mini); } printf("%d %d %d\n",sum,mmin,mmax); } } } }