HDU 1698 Just a Hook【线段树】题解

系统管理员 2021-07-24 18:04 359阅读 0赞

### 目录 ###

*   *   *  1.题目
         *  2.代码

### 1.题目 ###

In the game of DotA, Pudge’s meat hook is actually the most horrible thing for most of the heroes. The hook is made up of several consecutive metallic sticks which are of the same length.

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Now Pudge wants to do some operations on the hook.

Let us number the consecutive metallic sticks of the hook from 1 to N. For each operation, Pudge can change the consecutive metallic sticks, numbered from X to Y, into cupreous sticks, silver sticks or golden sticks.  
The total value of the hook is calculated as the sum of values of N metallic sticks. More precisely, the value for each kind of stick is calculated as follows:

For each cupreous stick, the value is 1.  
For each silver stick, the value is 2.  
For each golden stick, the value is 3.

Pudge wants to know the total value of the hook after performing the operations.  
You may consider the original hook is made up of cupreous sticks.  
Input  
The input consists of several test cases. The first line of the input is the number of the cases. There are no more than 10 cases.  
For each case, the first line contains an integer N, 1<=N<=100,000, which is the number of the sticks of Pudge’s meat hook and the second line contains an integer Q, 0<=Q<=100,000, which is the number of the operations.  
Next Q lines, each line contains three integers X, Y, 1<=X<=Y<=N, Z, 1<=Z<=3, which defines an operation: change the sticks numbered from X to Y into the metal kind Z, where Z=1 represents the cupreous kind, Z=2 represents the silver kind and Z=3 represents the golden kind.  
Output  
For each case, print a number in a line representing the total value of the hook after the operations. Use the format in the example.  
Sample Input

1
    10
    2
    1 5 2
    5 9 3

Sample Output

Case 1: The total value of the hook is 24

### 2.代码 ###

#define _CRT_SECURE_NO_WARNINGS 
    #include<iostream>
    #include<cstdio>
    #include<algorithm>
    using namespace std;
    const int maxn = 1e5 + 10;
    #define lson l,m,rt<<1 //宏定义
    #define rson m+1,r,rt<<1|1 
    int col[maxn << 2]; //懒惰标记
    int sum[maxn << 2]; //区间的和
    void PushUP(int rt)///更新节点
    { 
    	sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
    }
    void PushDown(int rt, int m)///下推标记 m表示左右子树总的数量
    { 
    	if (col[rt]) ///有标记 
    	{ 
    		col[rt << 1] = col [rt << 1 | 1] = col[rt];//下推
    		sum[rt << 1] = (m - (m >> 1))*col[rt];
    		sum[rt << 1 | 1] = (m >> 1)*col[rt];
    		col[rt] = 0;  ///清除标记
    	}
    }
    void build(int l, int r, int rt)//建树
    { 
    	col[rt] = 0;
    	sum[rt]= 1;  ///初始值均为1
    	if (l == r) return;//到达叶节点 结束
    	int m = (l + r) >> 1;
    	build(lson); //左右递归
    	build(rson);
    	PushUP(rt);  ///更新节点
    }
    void update(int L, int R,int c, int l, int r, int rt) ///区间更新
    { 
    	if (L <= l && r <= R)
    	{ 
    		col[rt] = c;  懒惰标记
    		sum[rt] = c * (r - l + 1);
    		return;
    	}
    	PushDown(rt, r - l + 1); //x下推标记
    	int m = (l + r) >> 1;
    	if (L <= m)  update(L, R, c, lson);
    	if (R > m)   update(L, R, c, rson);
    	PushUP(rt);
    }
    int main()
    { 
    	int T, n, m;
    	scanf("%d", &T);
    	for (int cas = 1; cas <= T; cas++)
    	{ 
    		scanf("%d%d", &n, &m);
    		build(1, n, 1);
    		while (m--)
    		{ 
    			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", cas, sum[1]);
    	}
    	return 0;
    }

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