poj 3468 A Simple Problem with Integers 线段树 题解《挑战程序设计竞赛》

地址 http://poj.org/problem?id=3468

线段树模板

要背下此模板

线段树

 #include <iostream>
 #include <vector>
 #include <math.h>
 #include <algorithm>

 using namespace std;

 /*
 Sample Input
 10 5
 1 2 3 4 5 6 7 8 9 10
 Q 4 4
 Q 1 10
 Q 2 4
 C 3 6 3
 Q 2 4
 Sample Output
 4
 55
 9
 15
 */
 typedef long long ll;

 const int  DAT_SIZE = ( << ) - ;

 const int MAX_N = ;

 //输入
 int N, Q;
 int A[MAX_N];
 char T[MAX_N];
 int L[MAX_N], R[MAX_N], X[MAX_N];

 //线段树
 ll dat_a[DAT_SIZE], dat_b[DAT_SIZE];

 //对区间[a,b]同时加x
 //k是节点的编号 对应的空间是[l,r)
 void add(int a, int b, int x, int k, int l, int r)
 {
     if (a <= l && r <= b) {
         dat_a[k] += x;
     }
     else if (l < b && a < r) {
         dat_b[k] += (min(b,r)-max(a,l))* x;
         add(a, b, x, k *  + , l, (l + r) / );
         add(a, b, x, k *  + , (l + r) / , r);
     }
 }

 //计算[a,b)的和
 //k是节点的编号 对应的区间是[l,r)
 ll sum(int a, int b, int k, int l, int r)
 {
     if (b <= l || r <= a) {
         return ;
     }
     else if(a <= l && r <= b){
         return dat_a[k] * (r - l) + dat_b[k];
     }
     else {
         ll res = (min(b,r)-max(a,l)) * dat_a[k];
         res += sum(a, b, k *  + , l, (l + r) / );
         res += sum(a, b, k *  + , (l + r) / , r);
         return res;
     }
 }

 void solve()
 {
     for (int i = ; i < N; i++) {
         add(i, i + , A[i], , , N);
     }
     for (int i = ; i < Q; i++) {
         if (T[i] == 'C') {
             add(L[i]-, R[i] , X[i], , , N);
         }
         else {
             printf("%lld\n",sum(L[i]-,R[i],,,N));
         }
     }
 }

 int main()
 {
     cin >> N >> Q;

     for (int i = ; i < N; i++) {
         cin >> A[i];
     }

     for (int i = ; i < Q; i++) {
         cin >> T[i];
         if (T[i] == 'C')
             cin >> L[i] >> R[i] >> X[i];
         else
             cin >> L[i] >> R[i];
     }

     solve();

     return ;
 }

 #include <iostream>
 #include <vector>

 using namespace std;

 /*
 Sample Input
 10 5
 1 2 3 4 5 6 7 8 9 10
 Q 4 4
 Q 1 10
 Q 2 4
 C 3 6 3
 Q 2 4
 Sample Output
 4
 55
 9
 15
 */

 typedef long long ll;

 int n, m;

 const int MAX_N  = ;
 int input[MAX_N];

 struct Node {
     int l, r;
     ll data;
 };

 struct Node stree[MAX_N * ];

 void build(int idx, int l, int r)
 {
     stree[idx].l = l;
     stree[idx].r = r;
     if (l == r) {
         //叶子节点
         stree[idx].data = input[l];
         return;
     }

     int mid = (l + r) / ;
     build(idx * , l,mid);
     build(idx *  + , mid + , r);

     stree[idx].data = stree[idx * ].data + stree[idx *  + ].data;
 }

 ll query(int idx, int start, int end, int l, int r)
 {
     if (l == stree[idx].l && r == stree[idx].r) {
         return stree[idx].data;
     }

     int mid = (start + end) / ;
     if (l <= mid && r <= mid) {
         return query(idx * , start, mid, l, r);
     }
     else if (l > mid && r > mid) {
         return query(idx *  + , mid + , end, l, r);
     }
     else {
         ll res = ;
         res += query(idx *  , start, mid, l, mid);
         res += query(idx *  + , mid + , end, mid + , r);
         return res;
     }

 }

 void add(int idx, int l, int r, int v)
 {
     if (l == r && stree[idx].l == stree[idx].r && stree[idx].l == l) {
         //叶子节点
         stree[idx].data += v;
         return;
     }

     int mid = (stree[idx].l + stree[idx].r) / ;
     if (l <= mid && r <= mid) {
         add(idx * , l, r,v);
     }
     else if (l > mid && r > mid) {
         add(idx *  + , l, r,v);
     }
     else {
         add(idx *  , l, mid, v);
         add(idx *  + , mid+, r, v);
     }

     stree[idx].data = stree[idx*].data + stree[idx*+].data;
 }

 int main()
 {
     int n, m;
     cin >> n >> m;
     for (int i = ; i <= n; i++) {
         cin >> input[i];
     }

     build(, , n);

     for (int i = ; i < m; i++) {
         char t;
         int l, r, v;
         cin >> t;
         if (t == 'C') {
             cin >> l >> r >> v;
             add(, l, r, v);
         }
         else if (t == 'Q') {
             cin >> l >> r;
             cout << query(, , n, l, r) << endl;
         }
     }

     return ;
 }

自写TLE代码