HDU

题意：n个棍子，初始值全为1，给定Q个区间，分别赋值，问n个棍子的总值。

分析：lazy标记主要体现在update上。

当l <= L && R <= r时，该结点的子结点值不再更新，取而代之的是给该结点一个lazy值，以记录下来该结点的子结点并没有更新。

当赋值的区间落在子结点上时，才将lazy标记传递，同时更新子结点相应的sum值。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 100000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int sum[MAXN << 2];
int lazy[MAXN << 2];
void build(int id, int L, int R, int v){
    if(L == R){
        sum[id] = v;
    }
    else{
        int mid = L + (R - L) / 2;
        build(id << 1, L, mid, v);
        build(id << 1 | 1, mid + 1, R, v);
        sum[id] = sum[id << 1] + sum[id << 1 | 1];
    }
}
void pushdown(int id, int L, int R){
    if(lazy[id]){
        lazy[id << 1] = lazy[id << 1 | 1] = lazy[id];
        int mid = L + (R - L) / 2;
        sum[id << 1] = (mid - L + 1) * lazy[id];
        sum[id << 1 | 1] = (R - mid) * lazy[id];
        lazy[id] = 0;
    }
}
void update(int l, int r, int id, int L, int R, int v){
    if(l <= L && R <= r){
        sum[id] = (R - L + 1) * v;
        lazy[id] = v;
    }
    else{
        pushdown(id, L, R);
        int mid = L + (R - L) / 2;
        if(l <= mid) update(l, r, id << 1, L, mid, v);
        if(r > mid) update(l, r, id << 1 | 1, mid + 1, R, v);
        sum[id] = sum[id << 1] + sum[id << 1 | 1];
    }
}
int main(){
    int T;
    scanf("%d", &T);
    int kase = 0;
    while(T--){
        memset(sum, 0, sizeof sum);
        memset(lazy, 0, sizeof lazy);
        int N;
        scanf("%d", &N);
        build(1, 1, N, 1);
        int Q;
        scanf("%d", &Q);
        while(Q--){
            int X, Y, Z;
            scanf("%d%d%d", &X, &Y, &Z);
            update(X, Y, 1, 1, N, Z);
        }
        printf("Case %d: The total value of the hook is %d.
", ++kase, sum[1]);
    }
    return 0;
}