题意：有一个 r*c 的全 0矩阵, 进行 3 种操作。
1 x1 y1 x2 y2 val 表示将(x1,y1,x2,y2)（x1<=x2,y1<=y2）子矩阵中的所有元素加val；
2 x1 y1 x2 y2 val 表示将(x1,y1,x2,y2)（x1<=x2,y1<=y2）子矩阵中的所有元素变为val；
3 x1 y1 x2 y2 val 表示输出(x1,y1,x2,y2)（x1<=x2,y1<=y2）子矩阵中的所有元素的和，最小值和最大值。
共有m次操作(1<=m<=20000) 。 矩阵不超过20 行，元素总数不超过 1e6 。
tags：因为不超过 20行，所以我们每行建棵线段树，然后就是基本的区间更新查询。
#include<bits/stdc++.h> using namespace std; #pragma comment(linker, "/STACK:102400000,102400000") #define rep(i,a,b) for (int i=a; i<=b; ++i) #define per(i,b,a) for (int i=b; i>=a; --i) #define mes(a,b) memset(a,b,sizeof(a)) #define INF 0x3f3f3f3f #define MP make_pair #define PB push_back #define fi first #define se second #define mid (l+(r-l)/2) #define PIII pair<ll, pair< ll, ll > > typedef long long ll; const int N = 1000005; struct Item { ll sum, minn, maxn, addv, setv; } ; vector< Item > tr[21]; int n, m, q; void Init() { rep(i,1,n) tr[i].clear(); rep(i,1,n) rep(j,1,m<<2) tr[i].PB(Item()); } void pushdown(int ii, int ro, int l, int r) { if(tr[ii][ro].setv) { tr[ii][ro<<1].addv = tr[ii][ro<<1|1].addv = 0; tr[ii][ro<<1].setv = tr[ii][ro<<1|1].setv = tr[ii][ro].setv; tr[ii][ro<<1].sum = tr[ii][ro].setv*(mid-l+1); tr[ii][ro<<1|1].sum = tr[ii][ro].setv*(r-mid); tr[ii][ro<<1].maxn = tr[ii][ro<<1].minn = tr[ii][ro].setv; tr[ii][ro<<1|1].maxn = tr[ii][ro<<1|1].minn = tr[ii][ro].setv; } tr[ii][ro<<1].addv += tr[ii][ro].addv; tr[ii][ro<<1|1].addv += tr[ii][ro].addv; tr[ii][ro<<1].sum += tr[ii][ro].addv*(mid-l+1); tr[ii][ro<<1|1].sum += tr[ii][ro].addv*(r-mid); tr[ii][ro<<1].maxn += tr[ii][ro].addv; tr[ii][ro<<1].minn += tr[ii][ro].addv; tr[ii][ro<<1|1].maxn += tr[ii][ro].addv; tr[ii][ro<<1|1].minn += tr[ii][ro].addv; tr[ii][ro].addv = tr[ii][ro].setv = 0; } void pushup(int ii, int ro) { tr[ii][ro].maxn = max(tr[ii][ro<<1].maxn, tr[ii][ro<<1|1].maxn); tr[ii][ro].minn = min(tr[ii][ro<<1].minn, tr[ii][ro<<1|1].minn); tr[ii][ro].sum = tr[ii][ro<<1].sum + tr[ii][ro<<1|1].sum; } void update(int ii, int ro, int l, int r, int ql, int qr, int flag, int vi) { if(ql<=l && r<=qr) { if(flag==1) { tr[ii][ro].addv += vi; tr[ii][ro].sum += 1LL*vi*(r-l+1); tr[ii][ro].maxn += vi; tr[ii][ro].minn += vi; } else { tr[ii][ro].setv = vi; tr[ii][ro].addv = 0; tr[ii][ro].sum = 1LL*vi*(r-l+1); tr[ii][ro].maxn = tr[ii][ro].minn = vi; } return ; } pushdown(ii, ro, l, r); if(mid<qr) update(ii, ro<<1|1, mid+1, r, ql, qr, flag, vi); if(ql<=mid) update(ii, ro<<1, l, mid, ql, qr, flag, vi); pushup(ii, ro); } PIII get_ans(PIII x, PIII y) { PIII ret; ret.fi = x.fi+y.fi; ret.se.fi = max(x.se.fi, y.se.fi); ret.se.se = min(x.se.se, y.se.se); return ret; } PIII query(int ii, int ro, int l, int r, int ql, int qr) { if(ql<=l && r<=qr) { return MP(tr[ii][ro].sum, MP(tr[ii][ro].maxn, tr[ii][ro].minn)); } pushdown(ii, ro, l, r); PIII ret = MP(0, MP(0,1e18)); if(mid<qr) ret = get_ans(ret, query(ii, ro<<1|1, mid+1, r, ql, qr)); if(ql<=mid) ret = get_ans(ret, query(ii, ro<<1, l, mid, ql, qr)); pushup(ii, ro); return ret; } int main() { while(~scanf("%d%d%d", &n, &m, &q)) { Init(); int ti, x1, y1, x2, y2, vi; while(q--) { scanf("%d", &ti); if(ti==1) { scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &vi); rep(i,x1,x2) update(i, 1, 1, m, y1, y2, 1, vi); } else if(ti==2) { scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &vi); rep(i,x1,x2) update(i, 1, 1, m, y1, y2, 2, vi); } else { scanf("%d%d%d%d", &x1, &y1, &x2, &y2); PIII ans = MP(0, MP(0,1e18)); rep(i,x1,x2) ans = get_ans(ans, query(i, 1, 1, m, y1, y2)); printf("%lld %lld %lld\n", ans.fi, ans.se.se, ans.se.fi); } } } return 0; }
