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*];
     Tree(){
         memset(f, , sizeof(f));
     }
     void init(){
         memset(f, , sizeof(f));
     }
     void push_down(int c){
         int l = lson(c), r = rson(c);
         f[l] += f[c];
         f[r] += f[c];
         f[c] = ;
     }
     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 + , r, rson(c), lp, rp, d);
         else{
             update(l, m, lson(c), lp, m, d);
             update(m+, r, rson(c), m+, 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+, 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 = ; i <= n; i ++){
             scanf("%I64d", &a[i]);
         }
         for(int i = ; i <= m; i ++){
             scanf("%d %d %I64d", &ls[i], &rs[i], &ind[i]);
         }
         for(int i = , x, y; i <= k; i ++){
             scanf("%d%d", &x, &y);
             insTree.update(, m, , x, y, );
         }
         insTree.query(, , m, ind, );
         for(int i = ; i <= m; i ++){
             arrTree.update(, n, , ls[i], rs[i], ind[i]);
         }
         arrTree.query(, , n, a, );
         for(int i = ; 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, , sizeof(lx));
         memset(ly, , sizeof(ly));
         for(int i = ; i <= n; i ++) scanf("%I64d", &a[i]);
         for(int i = ; i <= m; i ++) scanf("%d%d%I64d", &px[i], &py[i], &ind[i]);
         for(int i = , x, y; i <= k; i ++){
             scanf("%d%d", &x, &y); lx[x] += , lx[y+] -= ;
         }
         LL s = ;
         for(int i = ; i <= m; i ++){
             s += lx[i];
             ind[i] = ind[i] *s;
         }
         for(int i = ; i <= m; i ++){
             ly[px[i]] += ind[i];
             ly[py[i]+] -= ind[i];
         }
         s = ;
         for(int i = ; i <= n; i ++){
             s += ly[i];
             printf("%I64d ", a[i] + s);
         }
         printf("\n");
     }
     return ;
 }