标签：int opera define 直线 nbsp isp 打表 两种 max
给定一个$6\times n$的方格，每个点有一个非负权值，有两种操作
$n\le 10^5$
打表枚举观察可得，最终路径由最多$3$步组成
所以我们需要记录一些$6\times6$的矩阵，令当前处理区间为$[l,r]$，区间中点为$mid$：
强调：所有路径均为左开右闭
区间信息的维护可以用线段树处理，考虑上传操作：
枚举$k\in[1,6]$，令$ls$为左儿子，$rs$为右儿子，不难推出
$$lm[i][j] = \min\{ls.lr[i][k]+rs.ll[k][j]+ls.rr[j][j]\}$$
$$rm[i][j]=\min\{ls.rr[i][k]+rs.lr[k][j]+rs.ll[j][j]\}$$
$$ll[i][j]=\min\{ls.ll[i][j],lm[i][k]+ls.lr[j][k]-ls.rr[k][k]\}$$
$$rr[i][j]=\min\{rs.rr[i][j],rm[k][i]+rs.lr[k][j]-rs.ll[k][k]\}$$
$$lr[i][j]=\min\{lm[i][k]+rm[k][j]-ls.rr[k][k]-rs.ll[k][k],ls.lr[i][k]+rs.lr[k][j]\}$$
可能的路径有$4$中情况（此处的$1,2,3$代表上面的三步）
令$ls=[1,y_1),mid=[y_1,y_2],rs=(y_2,n]$，枚举$i,j$
将$4$种情况的$ans$取最小值即可。
#include <bits/stdc++.h> #define inf 2139062143 using namespace std; int Read() { int x = 0, f = 1; char ch = getchar(); while (!isdigit(ch)) { if (ch == ‘-‘) f = -1; ch = getchar(); } while (isdigit(ch)) { x = (x << 3) + (x << 1) + ch - ‘0‘; ch = getchar(); } return x * f; } int n, a[7][100005]; struct node { int ll[7][7], rr[7][7], lr[7][7]; void Clear() { memset(ll, 0x7f, sizeof(ll)); memset(rr, 0x7f, sizeof(rr)); memset(lr, 0x7f, sizeof(lr)); } bool Empty() const { return ll[0][0] == inf; } } seg[400005]; node operator+(const node &ls, const node &rs) { int lm[7][7], rm[7][7]; if (ls.Empty()) return rs; if (rs.Empty()) return ls; node a; a.Clear(); memset(lm, 0x7f, sizeof(lm)); memset(rm, 0x7f, sizeof(rm)); for (int k = 0; k < 6; k++) { for (int i = 0; i < 6; i++) { for (int j = 0; j < 6; j++) { lm[i][j] = min(lm[i][j], ls.lr[i][k] + rs.ll[k][j] + ls.rr[j][j]); rm[i][j] = min(rm[i][j], rs.ll[i][i] + ls.rr[i][k] + rs.lr[k][j]); } } } for (int i = 0; i < 6; i++) { for (int j = 0; j < 6; j++) { a.ll[i][j] = ls.ll[i][j], a.rr[i][j] = rs.rr[i][j], a.lr[i][j] = inf; for (int k = 0; k < 6; k++) { a.ll[i][j] = min(a.ll[i][j], lm[i][k] + ls.lr[j][k] - ls.rr[k][k]); a.rr[i][j] = min(a.rr[i][j], rm[k][i] + rs.lr[k][j] - rs.ll[k][k]); a.lr[i][j] = min(a.lr[i][j], ls.lr[i][k] + rs.lr[k][j]); a.lr[i][j] = min(a.lr[i][j], lm[i][k] + rm[k][j] - ls.rr[k][k] - rs.ll[k][k]); } } } return a; } void Init(int o, int l) { int sum[7]; for (int i = 0; i < 6; i++) { sum[i] = (i ? sum[i - 1] : 0) + a[i][l]; } for (int i = 0; i < 6; i++) { for (int j = 0; j < 6; j++) { int tmp = sum[max(i, j)] - (min(j, i) ? sum[min(j, i) - 1] : 0); seg[o].ll[i][j] = seg[o].rr[i][j] = seg[o].lr[i][j] = tmp; } } } void build(int o, int l, int r) { if (l == r) { Init(o, l); return; } int mid = (l + r) >> 1; build(o << 1, l, mid); build(o << 1 | 1, mid + 1, r); seg[o] = seg[o << 1] + seg[o << 1 | 1]; } void modify(int o, int l, int r, int pos) { if (l == r) { Init(o, l); return; } int mid = (l + r) >> 1; if (pos <= mid) modify(o << 1, l, mid, pos); else modify(o << 1 | 1, mid + 1, r, pos); seg[o] = seg[o << 1] + seg[o << 1 | 1]; } node query(int o, int l, int r, int nl, int nr) { if (nl <= l && r <= nr) { return seg[o]; } node ans; ans.Clear(); int mid = (l + r) >> 1; if (nl <= mid) ans = ans + query(o << 1, l, mid, nl, nr); if (mid < nr) ans = ans + query(o << 1 | 1, mid + 1, r, nl, nr); return ans; } int Qmin(int sx, int sy, int tx, int ty) { node ls = query(1, 1, n, 1, sy); node md = query(1, 1, n, sy, ty); node rs = query(1, 1, n, ty, n); int ans = md.lr[sx][tx]; for (int i = 0; i < 6; i++) { for (int j = 0; j < 6; j++) { ans = min(ans, md.ll[sx][i] + ls.rr[i][j] + md.lr[j][tx] - a[i][sy] - a[j][sy]); ans = min(ans, md.lr[sx][i] + rs.ll[i][j] + md.rr[j][tx] - a[i][ty] - a[j][ty]); ans = min(ans, ls.rr[sx][i] + md.lr[i][j] + rs.ll[j][tx] - a[i][sy] - a[j][ty]); } } return ans; } int main() { n = Read(); for (int i = 0; i < 6; i++) { for (int j = 1; j <= n; j++) { a[i][j] = Read(); } } build(1, 1, n); int m = Read(); for (int i = 1; i <= m; i++) { int opt = Read(); if (opt == 1) { int x = Read(), y = Read(), z = Read(); a[x - 1][y] = z; modify(1, 1, n, y); } else { int x = Read(), y = Read(), z = Read(), w = Read(); if (y > w) swap(x, z), swap(y, w); printf("%d\n", Qmin(x - 1, y, z - 1, w)); } } }
标签：int opera define 直线 nbsp isp 打表 两种 max
原文地址：https://www.cnblogs.com/verjun/p/13385816.html
