hdu 1892二维树状数组

这题我知道是用树状数组,可是好久没打树状数组了,就想用普通方法水过去~~结果……结果……水了好多方法都水不过,出题人真狠呐……我的水方法是对于每一次查询,初始化ans=(x2-x1+1)*(y2-y1+1),然后对于这个操作之前的每一个操作,对ans进行处理即可。可是交上去TLE,加上输入外挂,还是TLE,又加一个优化,即对于每一个查询,如果查询的区间小于10000,就直接数,还是TLE,服了,还是打树状数组吧~~~~~~~~~~~~~


我的水代码:

/*
 * hdu1892/win.cpp
 * Created on: 2012-11-1
 * Author    : ben
 */
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
#include <stack>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <functional>
#include <numeric>
#include <cctype>
using namespace std;
const int MAXN = 1010;
typedef struct Ope{
    int x1, x2, y1, y2;
    int n;
    char type;
}Ope;
vector<Ope> ope;
int room[MAXN][MAXN];
inline int get_int() {
    int res = 0, ch;
    while (!((ch = getchar()) >= '0' && ch <= '9')) {
        if (ch == EOF)
            return 1 << 30;
    }
    res = ch - '0';
    while ((ch = getchar()) >= '0' && ch <= '9')
        res = res * 10 + (ch - '0');
    return res;
}
inline bool inrect(int &x, int &y, int &x1, int &y1, int &x2, int &y2) {
    return x >= x1 && x <= x2 && y >= y1 && y <= y2;
}
void work2(int x1, int y1, int x2, int y2, int n) {
    int ans = (x2 - x1 + 1) * (y2 - y1 + 1);
    for(int i = 0; i < n; i++) {
        if(ope[i].type == 'M') {
            if(inrect(ope[i].x1, ope[i].y1, x1, y1, x2, y2)) {
                ans -= ope[i].n;
            }
            if(inrect(ope[i].x2, ope[i].y2, x1, y1, x2, y2)) {
                ans += ope[i].n;
            }
        }else if(ope[i].type == 'A') {
            if(inrect(ope[i].x1, ope[i].y1, x1, y1, x2, y2)) {
                ans += ope[i].n;
            }
        }else if(ope[i].type == 'D') {
            if(inrect(ope[i].x1, ope[i].y1, x1, y1, x2, y2)) {
                ans -= ope[i].n;
            }
        }
    }
    printf("%d\n", ans);
}
void work1(Ope &o) {
    int ans = (o.x2 - o.x1 + 1) * (o.y2 - o.y1 + 1);
    if(ans < 1000) {
        ans = 0;
        for(int i = o.x1; i <= o.x2; i++) {
            for(int j = o.y1; j <= o.y2; j++) {
                ans += room[i][j];
            }
        }
        printf("%d\n", ans);
        o.type = 0;
    }
}
inline void input(Ope &o) {
    char c = getchar();
    int x1, y1, x2, y2;
    while(c <= ' ') {
        c = getchar();
    }
    o.type = c;
    if(c == 'S') {
        x1 = get_int(), y1 = get_int(), x2 = get_int(), y2 = get_int();
        o.x1 = min(x1, x2); o.x2 = max(x1, x2);
        o.y1 = min(y1, y2); o.y2 = max(y1, y2);
        work1(o);
    }else if(c == 'A') {
        o.x1 = get_int(), o.y1 = get_int(),    o.n = get_int();
        room[o.x1][o.y1] += o.n;
    }else if(c == 'D') {
        o.x1 = get_int(), o.y1 = get_int(), o.n = get_int();
        if(room[o.x1][o.y1] < o.n) {
            o.n = room[o.x1][o.y1];
        }
        room[o.x1][o.y1] -= o.n;
    }else  if(c == 'M') {
        o.x1 = get_int(), o.y1 = get_int(), o.x2 = get_int(), o.y2 = get_int();
        o.n = get_int();
        if(room[o.x1][o.y1] < o.n) {
            o.n = room[o.x1][o.y1];
        }
        room[o.x1][o.y1] -= o.n;
        room[o.x2][o.y2] += o.n;
    }
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
#endif
    int T, Q;
    T = get_int();
    for(int t = 1; t <= T; t++) {
        printf("Case %d:\n", t);
        Q = get_int();
        fill(*room, *room + MAXN * MAXN, 1);
        ope.resize(Q);
        for_each(ope.begin(), ope.end(), input);
        for(int i = 0, len = (int)ope.size(); i < len; i++) {
            if(ope[i].type == 'S') {
                work2(ope[i].x1, ope[i].y1, ope[i].x2, ope[i].y2, i);
            }
        }
    }
    return 0;
}

二维树状数组代码:

/*
 * hdu1892/win.cpp
 * Created on: 2011-9-6
 * Author    : ben
 */
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <map>
using namespace std;
typedef long long LL;
const int MAXN = 1005;
LL data[MAXN][MAXN];
int Row = 1001, Col = 1001;
inline int Lowbit(const int &x) { // x > 0
    return x & (-x);
}
LL sum(int x, int y) {
    LL ret = 0;
    for(int i = x; i > 0; i -= Lowbit(i)) {
        for(int j = y; j > 0; j -= Lowbit(j)) {
            ret += data[i][j];
        }
    }
    return ret;
}
void update(int x, int y, int delta) {
    for(int i = x; i <= Row; i += Lowbit(i)) {
        for(int j = y; j <= Col; j += Lowbit(j)) {
            data[i][j] += delta;
        }
    }
}
void work() {
    char c;
    int x1, y1, x2, y2, n;
    scanf(" %c", &c);
    if(c == 'S') {
        scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
        if(x1 > x2) {
            swap(x1, x2);
        }
        if(y1 > y2) {
            swap(y1, y2);
        }
        printf("%d\n", (int)(sum(x2 + 1, y2 + 1) - sum(x1, y2 + 1) + sum(x1, y1) - sum(x2 + 1, y1)));
    }else if(c == 'A') {
        scanf("%d%d%d", &x1, &y1, &n);
        update(x1 + 1, y1 + 1, n);
    }else if(c == 'D') {
        scanf("%d%d%d", &x1, &y1, &n);
        x1++, y1++;
        int t =sum(x1, y1) - sum(x1 - 1, y1) - sum(x1, y1 - 1) + sum(x1 - 1, y1 - 1);
        if(t < n) {
            n = t;
        }
        update(x1, y1, -n);
    }else if(c == 'M') {
        scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &n);
        x1++, y1++;
        x2++, y2++;
        int t =sum(x1, y1) - sum(x1 - 1, y1) - sum(x1, y1 - 1) + sum(x1 - 1, y1 - 1);
        if(t < n) {
            n = t;
        }
        update(x1, y1, -n);
        update(x2, y2, n);
    }
}
void init() {
    for(int i = 1; i <= Row; i++) {
        for(int j = 1; j <= Col; j++) {
            update(i, j, 1);
        }
    }
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
#endif
    int T, Q;
    scanf("%d", &T);
    for(int t = 1; t <= T; t++) {
        printf("Case %d:\n", t);
        scanf("%d", &Q);
        fill(*data, *data + MAXN * MAXN, 0);
        init();
        while(Q--) {
            work();
        }
    }
    return 0;
}
原文地址:https://www.cnblogs.com/moonbay/p/2749579.html