bzoj3995
线段树
额 计蒜客竟然把这个出成noip模拟题。。。
这个东西很像1018，只不过维护的东西不太一样
具体有这五种情况，合并请看代码，自己写了一个结果wa了，然后就copy了一下。。。
然后build的时候不用把叶子结点的值赋成inf，感觉奥妙重重
#include<bits/stdc++.h> using namespace std; const int N = 60010; int n, m; int tree[N << 3][5], a[N], b[N], c[N], ans[5]; void up(int &a, const int &b) { a = min(a, b); } /* void maintain(int a[5], int l[5], int r[5], int mn, int sm) { memset(a, 0x3f3f3f3f, sizeof(tree[0])); up(a[0], l[0] + r[0] + mn); up(a[0], l[0] + r[3] + sm); up(a[0], l[3] + r[0] + sm); up(a[0], l[1] + r[0] + sm); up(a[0], l[0] + r[2] + sm); up(a[1], l[0] + r[1] + mn); up(a[1], l[3] + r[1] + sm); up(a[1], l[0] + r[4] + sm); up(a[1], l[0] + r[3] + mn); up(a[1], l[0] + r[4] + mn); up(a[1], l[0] + r[2] + mn); up(a[2], l[2] + r[0] + mn); up(a[2], l[4] + r[0] + sm); up(a[2], l[3] + a[0] + mn); up(a[2], l[1] + r[0] + mn); up(a[2], l[2] + r[3] + sm); up(a[2], l[2] + r[2] + sm); up(a[3], l[3] + r[3] + sm); up(a[4], l[2] + r[1] + mn); up(a[4], l[2] + r[3] + mn); up(a[4], l[2] + r[4] + sm); up(a[4], l[3] + r[1] + mn); up(a[4], l[3] + r[4] + sm); up(a[4], l[2] + r[1] + mn); up(a[4], l[4] + r[3] + sm); up(a[4], l[4] + r[1] + sm); up(a[4], l[3] + r[3] + mn); } */ void maintain(int a[5] , int l[5] , int r[5] , int mn , int sm) { memset(a , 0x3f3f , sizeof(tree[0])); up(a[0] , l[0] + r[0] + mn); up(a[1] , l[0] + r[1] + mn); up(a[0] , l[0] + r[2] + sm); up(a[0] , l[0] + r[3] + sm); up(a[1] , l[0] + r[3] + mn); up(a[1] , l[0] + r[4] + sm); up(a[0] , l[1] + r[0] + sm); up(a[1] , l[1] + r[1] + sm); up(a[1] , l[1] + r[3] + sm); up(a[2] , l[2] + r[0] + mn); up(a[4] , l[2] + r[1] + mn); up(a[2] , l[2] + r[2] + sm); up(a[2] , l[2] + r[3] + sm); up(a[4] , l[2] + r[3] + mn); up(a[4] , l[2] + r[4] + sm); up(a[0] , l[3] + r[0] + sm); up(a[1] , l[3] + r[1] + sm); up(a[4] , l[3] + r[1] + mn); up(a[3] , l[3] + r[3] + sm); up(a[4] , l[3] + r[3] + mn); up(a[4] , l[3] + r[4] + sm); up(a[2] , l[4] + r[0] + sm); up(a[4] , l[4] + r[1] + sm); up(a[4] , l[4] + r[3] + sm); } void build(int l, int r, int o) { if(l == r) { tree[o][0] = c[l]; // tree[o][1] = tree[o][2] = 0x3f3f3f3f; return; } int mid = (l + r) >> 1; build(l, mid, o << 1); build(mid + 1, r, o << 1 | 1); int mn = min(a[mid], b[mid]), sm = a[mid] + b[mid]; maintain(tree[o], tree[o << 1], tree[o << 1 | 1], mn, sm); // printf("l = %d r = %d\n", l, r); // for(int i = 0; i < 5; ++i) printf("tree[%d][%d] = %d\n", o, i, tree[o][i]); } void update(int l, int r, int o, int pos) { if(l == r) { tree[o][0] = c[l]; return; } int mid = (l + r) >> 1; if(pos <= mid) update(l, mid, o << 1, pos); else update(mid + 1, r, o << 1 | 1, pos); int mn = min(a[mid], b[mid]), sm = a[mid] + b[mid]; maintain(tree[o], tree[o << 1], tree[o << 1 | 1], mn, sm); } bool query(int l, int r, int o, int la, int lb, int tmp[5]) { if(l > lb || r < la) return false; if(l >= la && r <= lb) { for(int i = 0; i < 5; ++i) tmp[i] = tree[o][i]; return true; } int mid = (l + r) >> 1, tmp1[5], tmp2[5]; bool flag1 = query(l, mid, o << 1, la, lb, tmp1); bool flag2 = query(mid + 1, r, o << 1 | 1, la, lb, tmp2); if(flag1 && flag2) { int mn = min(a[mid], b[mid]), sm = a[mid] + b[mid]; maintain(tmp, tmp1, tmp2, mn, sm); return true; } if(flag1) { for(int i = 0; i < 5; ++i) tmp[i] = tmp1[i]; return true; } if(flag2) { for(int i = 0; i < 5; ++i) tmp[i] = tmp2[i]; return true; } return false; } int main() { cin >> n >> m; for(int i = 1; i < n; ++i) scanf("%d", &a[i]); for(int i = 1; i < n; ++i) scanf("%d", &b[i]); for(int i = 1; i <= n; ++i) scanf("%d", &c[i]); build(1, n, 1); while(m--) { int x0, x1, y0, y1, w; char opt[10]; scanf("%s", opt); if(opt[0] == 'C') { scanf("%d%d%d%d%d", &x0, &y0, &x1, &y1, &w); if(x0 == x1) { if(y0 > y1) swap(y0, y1); if(x0 == 1) a[y0] = w; if(x0 == 2) b[y0] = w; update(1, n, 1, y0); update(1, n, 1, y1); } if(y0 == y1) { c[y0] = w; update(1, n, 1, y0); } } if(opt[0] == 'Q') { scanf("%d%d", &y0, &y1); query(1, n, 1, y0, y1, ans); printf("%d\n", ans[0]); } } return 0; }
