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题意：对一个矩阵进行子矩阵操作。
元素对多有1e6个，树套树不好开（我不会），把二维坐标化成一维的，一个子矩阵操作分解成多条线段的操作。
一次操作的复杂度是RlogC，很容易找到极端的数据（OJ上实测没有），如果判断一下然后建树复杂度是min(RlogC,ClogR)。
代码中结点没有保存l和r，而且询问是保存在全局变量中，这样做比较省空间。但是也有缺点，比如推区间结点数量的时候会麻烦一点。
#include<bits/stdc++.h> using namespace std; const int maxn = 1e6+1; int R,C; #define lid (id<<1) #define rid (id<<1|1) struct Seg { int add,setv; int Max,Min,sum; }tr[maxn<<2]; #define OP1(id,val)\ tr[id].add += val; tr[id].Max += val; tr[id].Min += val; tr[id].sum += (r-l+1)*val; #define OP2(id,val)\ tr[id].Max = tr[id].setv = tr[id].Min = val; tr[id].add = 0; tr[id].sum = val*(r-l+1); inline void push_down(int id,int l,int r) { int lc = lid, rc = rid, mid = (l+r)>>1; if(tr[id].setv>=0){ int &t = tr[id].setv; swap(r,mid); OP2(lc,t); swap(l,r); l++; swap(mid,r); OP2(rc,t); // l--; swap(mid,l); t = -1; } if(tr[id].add>0){ int &t = tr[id].add; swap(r,mid); OP1(lc,t); swap(l,r); l++; swap(mid,r); OP1(rc,t); // l--; swap(mid,l); t = 0; } } inline void maintain(int id) { int lc = lid, rc = rid; tr[id].sum = tr[lc].sum + tr[rc].sum; tr[id].Max = max(tr[lc].Max,tr[rc].Max); tr[id].Min = min(tr[lc].Min,tr[rc].Min); } int ql,qr,val; void add1D(int l = 0,int r = R*C-1,int id = 1) { if(ql<=l&&r<=qr) { OP1(id,val) return; } int mid = (l+r)>>1, lc = lid, rc = rid; push_down(id,l,r); if(ql<=mid) add1D(l,mid,lc); if(qr>mid) add1D(mid+1,r,rc); maintain(id); } void set1D(int l = 0,int r = R*C-1,int id = 1) { if(ql<=l&&r<=qr) { OP2(id,val) return; } int mid = (l+r)>>1, lc = lid, rc = rid; push_down(id,l,r); if(ql<=mid) set1D(l,mid,lc); if(qr>mid) set1D(mid+1,r,rc); maintain(id); } int queryMax1D(int l = 0,int r = R*C-1,int id = 1) { if(ql<=l&&r<=qr) { return tr[id].Max; } int mid = (l+r)>>1, lc = lid, rc = rid; push_down(id,l,r); int ret = 0; if(ql<=mid) ret = max(ret,queryMax1D(l,mid,lc)); if(qr>mid) ret = max(ret,queryMax1D(mid+1,r,rc)); return ret; } const int INF = 0x3f3f3f3f; int queryMin1D(int l = 0,int r = R*C-1,int id = 1) { if(ql<=l&&r<=qr) { return tr[id].Min; } int mid = (l+r)>>1, lc = lid, rc = rid; push_down(id,l,r); int ret = INF; if(ql<=mid) ret = min(ret,queryMin1D(l,mid,lc)); if(qr>mid) ret = min(ret,queryMin1D(mid+1,r,rc)); return ret; } int querySum1D(int l = 0,int r = R*C-1,int id = 1) { if(ql<=l&&r<=qr) { return tr[id].sum; } int mid = (l+r)>>1, lc = lid, rc = rid; push_down(id,l,r); int ret = 0; if(ql<=mid) ret += querySum1D(l,mid,lc); if(qr>mid) ret += querySum1D(mid+1,r,rc); return ret; } //[0,r) void add2D(int x1,int y1,int x2,int y2,int v) { val = v; int st = x1*C+y1, len = y2-y1; for(int x = x1; x <= x2; x++){ ql = st; qr = st+len; add1D(); st += C; } } void set2D(int x1,int y1,int x2,int y2,int v) { val = v; int st = x1*C+y1, len = y2-y1; for(int x = x1; x <= x2; x++){ ql = st; qr = st+len; set1D(); st += C; } } int querySum2D(int x1,int y1,int x2,int y2) { int ret = 0; int st = x1*C+y1, len = y2-y1; for(int x = x1; x <= x2; x++){ ql = st; qr = st+len; ret += querySum1D(); st += C; } return ret; } int queryMax2D(int x1,int y1,int x2,int y2) { int ret = 0; int st = x1*C+y1, len = y2-y1; for(int x = x1; x <= x2; x++){ ql = st; qr = st+len; ret = max(ret,queryMax1D()); st += C; } return ret; } int queryMin2D(int x1,int y1,int x2,int y2) { int ret = INF; int st = x1*C+y1, len = y2-y1; for(int x = x1; x <= x2; x++){ ql = st; qr = st+len; ret = min(ret,queryMin1D()); st += C; } return ret; } void int main() { //freopen("in.txt","r",stdin); int m; while(~scanf("%d%d%d",&R,&C,&m)){ ql = 0; qr = R*C-1; val = 0; set1D(); while(m--){ int op ,x1 ,y1 ,x2 ,y2 ; scanf("%d%d%d%d%d",&op,&x1,&y1,&x2,&y2); if(op == 1){ int v ; scanf("%d",&v); add2D(x1-1,y1-1,x2-1,y2-1,v); }else if(op == 2){ int v ; scanf("%d",&v); set2D(x1-1,y1-1,x2-1,y2-1,v); }else { x1--;x2--;y1--;y2--; printf("%d %d %d\n",querySum2D(x1,y1,x2,y2),queryMin2D(x1,y1,x2,y2),queryMax2D(x1,y1,x2,y2)); } } } return 0; }
UVA 11992 Fast Matrix Operations （降维）
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原文地址：http://www.cnblogs.com/jerryRey/p/4790224.html
