UVa 11297 Census (二维线段树)
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题意:给定上一个二维矩阵,有两种操作
第一种是修改 c x y val 把(x, y) 改成 val
第二种是查询 q x1 y1 x2 y2 查询这个矩形内的最大值和最小值。
析:二维线段树裸板。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #include <list> #include <assert.h> #include <bitset> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define fi first #define se second #define pb push_back #define sqr(x) ((x)*(x)) #define ms(a,b) memset(a, b, sizeof a) #define sz size() #define pu push_up #define pd push_down #define cl clear() #define all 1,n,1 #define FOR(x,n) for(int i = (x); i < (n); ++i) #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("in.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e17; const double inf = 1e20; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 500 + 10; const int mod = 1000; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c) { return r > 0 && r <= n && c > 0 && c <= m; } int minv[maxn<<2][maxn<<2]; int maxv[maxn<<2][maxn<<2]; int mmin, mmax; void push_upy(int x, int rt){ minv[x][rt] = min(minv[x][rt<<1], minv[x][rt<<1|1]); maxv[x][rt] = max(maxv[x][rt<<1], maxv[x][rt<<1|1]); } void push_upx(int rt, int x){ minv[rt][x] = min(minv[rt<<1][x], minv[rt<<1|1][x]); maxv[rt][x] = max(maxv[rt<<1][x], maxv[rt<<1|1][x]); } void buildy(int x, int l, int r, int rt, bool ok){ if(l == r){ if(ok){ scanf("%d", minv[x]+rt); maxv[x][rt] = minv[x][rt]; return ; } push_upx(x, rt); return ; } int m = l + r >> 1; buildy(x, lson, ok); buildy(x, rson, ok); push_upy(x, rt); } void buildx(int l, int r, int rt){ if(l == r){ buildy(rt, all, 1); return ; } int m = l + r >> 1; buildx(lson); buildx(rson); buildy(rt, all, 0); } void updatey(int x, int Y, int val, int l, int r, int rt, bool ok){ if(l == r){ if(ok){ minv[x][rt] = maxv[x][rt] = val; return ; } push_upx(x, rt); return ; } int m = l + r >> 1; if(Y <= m) updatey(x, Y, val, lson, ok); else updatey(x, Y, val, rson, ok); push_upy(x, rt); } void updatex(int X, int Y, int val, int l, int r, int rt){ if(l == r){ updatey(rt, Y, val, all, 1); return ; } int m = l + r >> 1; if(X <= m) updatex(X, Y, val, lson); else updatex(X, Y, val, rson); updatey(rt, Y, val, all, 0); } void queryy(int x, int L, int R, int l, int r, int rt){ if(L <= l && r <= R){ mmin = min(mmin, minv[x][rt]); mmax = max(mmax, maxv[x][rt]); return ; } int m = l + r >> 1; if(L <= m) queryy(x, L, R, lson); if(R > m) queryy(x, L, R, rson); } void queryx(int L, int R, int Y1, int Y2, int l, int r, int rt){ if(L <= l && r <= R){ queryy(rt, Y1, Y2, all); return ; } int m = l + r >> 1; if(L <= m) queryx(L, R, Y1, Y2, lson); if(R > m) queryx(L, R, Y1, Y2, rson); } int main(){ while(scanf("%d", &n) == 1){ buildx(1, n, 1); scanf("%d", &m); char op[5]; int x1, y1, x2 ,y2; while(m--){ scanf("%s %d %d %d", op, &x1, &y1, &x2); if(op[0] == ‘c‘) updatex(x1, y1, x2, all); else{ scanf("%d", &y2); mmin = INF; mmax = 0; queryx(x1, x2, y1, y2, all); printf("%d %d\n", mmax, mmin); } } } return 0; }
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