Codeforces 296C Greg and Array

数据结构题。个人认为是比较好的数据结构题。题意：给定一个长度为n的数组a，然后给定m个操作序列，每个操作：l, r, x将区间[l, r]内的元素都增加a,然后有k个查询，查询形式是对于操作序列x,y是将第x个操作到第y个操作执行一遍。然后求最后的数组的元素值。

1.线段树解法：维护两棵线段树，一棵用于维护执行的操作序列的执行次数，一棵用于维护数组a的值。复杂度O(nlogn)。

2.扫描区间。对于数组和操作序列分别维护一个数组lx[],ly[]。ly[i]表示区间[i, m]中每个操作执行的次数，lx[i]表示区间[i, n]中每个数的增量的值。O(n)的复杂度。

#include <stdio.h>
#include <string.h>
#define maxn 100005
#define lson(c) (c<<1)
#define rson(c) (c<<1|1)
#define mid(x, y) ((x+y)>>1)
typedef long long LL;

struct Tree{
    LL f[maxn*4];
    Tree(){
        memset(f, 0, sizeof(f));
    }
    void init(){
        memset(f, 0, sizeof(f));
    }
    void push_down(int c){
        int l = lson(c), r = rson(c);
        f[l] += f[c];
        f[r] += f[c];
        f[c] = 0;
    }
    void update(int l, int r, int c, int lp, int rp, LL d){
        if(lp <= l && r <= rp){
            f[c] += d;
            return ;
        }
        push_down(c);
        int m = mid(l, r);
        if(rp <= m) update(l, m, lson(c), lp, rp, d);
        else if(lp > m) update(m + 1, r, rson(c), lp, rp, d);
        else{
            update(l, m, lson(c), lp, m, d);
            update(m+1, r, rson(c), m+1, rp, d);
        }
    }
    void query(int c, int l, int r, LL a[], int s){
        if(l==r){
            if(s)
                a[l] = a[l]*f[c];
            else a[l] = a[l] + f[c];
            return ;
        }
        push_down(c);
        int mid = mid(l, r);
        query(lson(c), l, mid, a, s);
        query(rson(c), mid+1, r, a, s);
    }
}insTree, arrTree;
LL a[maxn], ind[maxn];
int ls[maxn], rs[maxn];

int main(){
    //freopen("test.in", "r", stdin);
    for(int n, m, k; scanf("%d%d%d", &n, &m, &k)!=EOF; ){
        insTree.init();
        arrTree.init();
        for(int i = 1; i <= n; i ++){
            scanf("%I64d", &a[i]);
        }
        for(int i = 1; i <= m; i ++){
            scanf("%d %d %I64d", &ls[i], &rs[i], &ind[i]);
        }
        for(int i = 1, x, y; i <= k; i ++){
            scanf("%d%d", &x, &y);
            insTree.update(1, m, 1, x, y, 1);
        }
        insTree.query(1, 1, m, ind, 1);
        for(int i = 1; i <= m; i ++){
            arrTree.update(1, n, 1, ls[i], rs[i], ind[i]);
        }
        arrTree.query(1, 1, n, a, 0);
        for(int i = 1; i <= n; i ++){
            printf("%I64d ", a[i]);
        }
        printf("\n");
    }
}

#include <stdio.h>
#include <string.h>
#define maxn 100005
typedef long long LL;
LL a[maxn], ind[maxn];
LL lx[maxn], ly[maxn];
int px[maxn], py[maxn];

int main(){
    //freopen("test.in", "r", stdin);
    for(int n, m, k; scanf("%d%d%d", &n, &m, &k)!=EOF; ){
        memset(lx, 0, sizeof(lx));
        memset(ly, 0, sizeof(ly));
        for(int i = 1; i <= n; i ++) scanf("%I64d", &a[i]);
        for(int i = 1; i <= m; i ++) scanf("%d%d%I64d", &px[i], &py[i], &ind[i]);
        for(int i = 1, x, y; i <= k; i ++){
            scanf("%d%d", &x, &y); lx[x] += 1, lx[y+1] -= 1;
        }
        LL s = 0;
        for(int i = 1; i <= m; i ++){
            s += lx[i];
            ind[i] = ind[i] *s;
        }
        for(int i = 1; i <= m; i ++){
            ly[px[i]] += ind[i];
            ly[py[i]+1] -= ind[i];
        }
        s = 0;
        for(int i = 1; i <= n; i ++){
            s += ly[i];
            printf("%I64d ", a[i] + s);
        }
        printf("\n");
    }
    return 0;
}