C. Census

Time Limit: 8 sec

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;
}

我要檢舉

#11297#Census

台長： Morris