题意：

屠夫的钩子区间是1~n，每段可能由铜，银，金组成，价值分别为1,2,3，进行一系列的更新之后，求钩子的总价值。

思路：

线段树的成段更新：要设置一个临时的线段树，每次更新的时候把更新段的值放在临时数组中，等到下次更新线段树的时候再向下更新（延迟更新）

 

#include 
    
     #define lhs l, m, rt << 1#define rhs m + 1, r, rt << 1 | 1const int maxn = 100010;int seg[maxn << 2];int col[maxn << 2];void PushUp(int rt){
         seg[rt] = seg[rt << 1] + seg[rt << 1 | 1];}void PushDown(int rt, int len){
         if (col[rt] > 0)    {
             col[rt << 1] = col[rt << 1 | 1] = col[rt];        seg[rt << 1] = (len - (len >> 1)) * col[rt];        seg[rt << 1 | 1] = (len >> 1) * col[rt];        col[rt] = 0;    }}void Build(int l, int r, int rt){
         col[rt] = 0;    if (l == r)        seg[rt] = 1;    else    {
             int m = (l + r) >> 1;        Build(lhs);        Build(rhs);        PushUp(rt);    }}void Update(int beg, int end, int value, int l, int r, int rt){
         if (beg <= l && r <= end)    {
             col[rt] = value;        seg[rt] = (r - l + 1) * value;    }    else    {
             PushDown(rt, r - l + 1);        int m = (l + r) >> 1;        if (beg <= m)            Update(beg, end, value, lhs);        if (end > m)            Update(beg, end, value, rhs);        PushUp(rt);    }}int main(){
         int cases, cnt = 0;    scanf("%d", &cases);    while (cases--)    {
             int n, q;        scanf("%d %d", &n, &q);        Build(1, n, 1);        for (int i = 0; i < q; ++i)        {
                 int a, b, c;            scanf("%d %d %d", &a, &b, &c);            Update(a, b, c, 1, n, 1);        }        printf("Case %d: The total value of the hook is %d.\n", ++cnt, seg[1]);    }    return 0;}