Just a Hook
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 23229 Accepted Submission(s): 11634
Problem Description
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.
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.
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.
题意： 给定一列数，初始的时候都是 1. 然后把分别把相应的 一段 i~ j 的值都改成2（或3）. 经过若干次更改， 求这列数的和。
分析： 建立线段树， 然后成段更新。 在更新线段树的值的同时， 算出父节点的sum值。
本题关键是PushDown函数的理解： 该函数的功能便是， 把线段数上的col[cur]的值（即： c）加到sum[]中，然后将col[]清零，以备下一次读取更改的c。
#include<cstdio>
#include<iostream>
using namespace std; #define lson l, m, cur<<1
#define rson m+1, r, cur<<1|1
const int maxn = ;
int col[maxn<<];
int sum[maxn<<]; void PushUp(int cur)
{
sum[cur] = sum[cur<<] + sum[cur<<|];
} void PushDown(int cur, int m)
{
if(col[cur])
{
col[cur<<] = col[cur<<|] = col[cur];
sum[cur<<] = (m-(m>>))*col[cur];
sum[cur<<|] = (m>>)*col[cur];
col[cur] = ;
}
} void Build(int l, int r, int cur)
{
col[cur] = ;
sum[cur] = ;
if(l==r) return;
int m = (r+l)>>;
Build(lson);
Build(rson);
PushUp(cur);
} void Update(int L, int R, int c, int l, int r, int cur)
{
if(L<=l&&r<=R)
{
col[cur] = c;
sum[cur] = c*(r-l+);
return;
}
PushDown(cur, r-l+);
int m = (l+r)>>;
if(L<=m) Update(L, R, c, lson);
if(R>m) Update(L, R, c, rson);
PushUp(cur);
} int main()
{
int T, n, m;
scanf("%d", &T);
for(int kase=; kase<=T; kase++)
{
scanf("%d%d", &n, &m);
Build(, n, );
while(m--)
{
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
Update(a, b, c, , n, );
}
printf("Case %d: The total value of the hook is %d.\n", kase, sum[]);
}
return ;
}