C. Census
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Time Limit: 8 sec
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Description
This year, there have been many problems with population
calculations, since in some cities, there are many emigrants, or the
population growth is very high. Every year the ACM (for Association for
Counting Members) conducts a census in each region. The country is
divided into N^2 regions, consisting of an N x N grid of regions.
Your task is to find the least, and the greatest population in some set
of regions. Since in a single year there is no significant change in
the populations, the ACM modifies the population counts by some number
of inhabitants.
The Input
In the first line you will find N (0 <= N <= 500), in following
the N lines you will be given N numbers, which represent, the initial
population of city C [i, j]. In the following line is the number Q (Q
<= 40000), followed by Q lines with queries:
There are two possible queries:
- "x1 y1 x2 y2" which represent the coordinates of the upper left and
lower right of where you must calculate the maximum and minimum change
in population.
- "x y v" indicating a change of the population of city C [x, y] by value v.
The Output
For each query, "x1 y1 x2 y2" print in a single line the greatest and
least amount of current population. Separated each output by a space.
Notice: There is only a single test case.
Sample Input |
Sample Output |
5 5 1 2 3 4 5 0 9 2 1 3 0 2 3 4 1 0 1 2 4 5 8 5 3 1 4 4 q 1 1 2 3 c 2 3 10 q 1 1 5 5 q 1 2 2 2 |
9 0 10 0 9 2 |
依照自己的想法去做實現, //C++ 0.648 Rank 33#include <stdio.h>#include <string.h>#define max(x, y) ((x) > (y) ? (x) : (y))#define min(x, y) ((x) < (y) ? (x) : (y))struct subTree { int l, r; int maxv, minv;};struct segmentTree { int l, r; subTree subtree[1025];} mainTree[1025];void subbuild(int fk, int nk, int l, int r) { mainTree[fk].subtree[nk].l = l; mainTree[fk].subtree[nk].r = r; if(l >= r) return; int m = (l+r)>>1; subbuild(fk, nk<<1, l, m); subbuild(fk, nk<<1|1, m+1, r);}void build(int k, int lr, int rr, int lc, int rc) { mainTree[k].l = lr; mainTree[k].r = rr; subbuild(k, 1, lc, rc); if(lr >= rr) return; int m = (lr+rr)>>1; build(k<<1, lr, m, lc, rc); build(k<<1|1, m+1, rr, lc, rc);}void submodify(int fk, int nk, int x, int y, int v) { if(mainTree[fk].subtree[nk].l == y && mainTree[fk].subtree[nk].r == y) { if(mainTree[fk].l == x && mainTree[fk].r == x) { mainTree[fk].subtree[nk].maxv = v; mainTree[fk].subtree[nk].minv = v; } else { mainTree[fk].subtree[nk].maxv = max(mainTree[fk<<1].subtree[nk].maxv, mainTree[fk<<1|1].subtree[nk].maxv); mainTree[fk].subtree[nk].minv = min(mainTree[fk<<1].subtree[nk].minv, mainTree[fk<<1|1].subtree[nk].minv); } } if(mainTree[fk].subtree[nk].l >= mainTree[fk].subtree[nk].r) return; int m = (mainTree[fk].subtree[nk].l+mainTree[fk].subtree[nk].r)>>1; if(y <= m) submodify(fk, nk<<1, x, y, v); else submodify(fk, (nk<<1)+1, x, y, v); if(mainTree[fk].l != mainTree[fk].r) { mainTree[fk].subtree[nk].maxv = max(mainTree[fk<<1].subtree[nk].maxv, mainTree[fk<<1|1].subtree[nk].maxv); mainTree[fk].subtree[nk].minv = min(mainTree[fk<<1].subtree[nk].minv, mainTree[fk<<1|1].subtree[nk].minv); } else { mainTree[fk].subtree[nk].maxv = max(mainTree[fk].subtree[nk<<1].maxv, mainTree[fk].subtree[nk<<1|1].maxv); mainTree[fk].subtree[nk].minv = min(mainTree[fk].subtree[nk<<1].minv, mainTree[fk].subtree[nk<<1|1].minv); }}void modify(int k, int x, int y, int v) { if(mainTree[k].l == x && mainTree[k].r == x) { submodify(k, 1, x, y, v); } if(mainTree[k].l >= mainTree[k].r) return; int m = (mainTree[k].l+mainTree[k].r)>>1; if(x <= m) modify(k<<1, x, y, v); else modify(k<<1|1, x, y, v); submodify(k, 1, x, y, v);}int ansMax, ansMin;void subquery(int fk, int nk, int lc, int rc) { if(mainTree[fk].subtree[nk].l == lc && mainTree[fk].subtree[nk].r == rc) { ansMax = max(ansMax, mainTree[fk].subtree[nk].maxv); ansMin = min(ansMin, mainTree[fk].subtree[nk].minv); return; } if(mainTree[fk].subtree[nk].l >= mainTree[fk].subtree[nk].r) return; int m = (mainTree[fk].subtree[nk].l+mainTree[fk].subtree[nk].r)>>1; if(rc <= m) subquery(fk, nk<<1, lc, rc); else if(lc > m) subquery(fk, nk<<1|1, lc, rc); else { subquery(fk, nk<<1, lc, m); subquery(fk, nk<<1|1, m+1, rc); }}void query(int k, int lr, int rr, int lc, int rc) { if(mainTree[k].l == lr && mainTree[k].r == rr) { subquery(k, 1, lc, rc); return; } if(mainTree[k].l >= mainTree[k].r) return; int m = (mainTree[k].l+mainTree[k].r)>>1; if(rr <= m) query(k<<1, lr, rr, lc, rc); else if(lr > m) query(k<<1|1, lr, rr, lc, rc); else { query(k<<1, lr, m, lc, rc); query(k<<1|1, m+1, rr,lc, rc); }}int main() { int n, m, q, x, y, v; int i, j, x1, x2, y1, y2; while(scanf("%d %d", &n, &m) == 2) { build(1, 1, n, 1, m); for(i = 1; i <= n; i++) { for(j = 1; j <= m; j++) { scanf("%d", &x); modify(1, i, j, x); } } scanf("%d", &q); char cmd[3]; while(q--) { scanf("%s", cmd); if(cmd[0] == 'q') { scanf("%d %d %d %d", &x1, &y1, &x2, &y2); ansMax = 0, ansMin = 0xfffffff; query(1, x1, x2, y1, y2); printf("%d %d\n", ansMax, ansMin); } else { scanf("%d %d %d %d", &x, &y, &v); modify(1, x, y, v); } } } return 0;}
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