Killing Monsters
Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 160????Accepted Submission(s): 86
Problem Description
Kingdom Rush is a popular TD game, in which you should build some towers to protect your kingdom from monsters. And now another wave of monsters is coming and you need again to know whether you can get through it.
The path of monsters is a straight line, and there are N blocks on it (numbered from 1 to N continuously). Before enemies come, you have M towers built. Each tower has an attack range [L, R], meaning that it can attack all enemies in every block i, where L<=i<=R. Once a monster steps into block i, every tower whose attack range include block i will attack the monster once and only once. For example, a tower with attack range [1, 3] will attack a monster three times if the monster is alive, one in block 1, another in block 2 and the last in block 3.
A witch helps your enemies and makes every monster has its own place of appearance (the ith monster appears at block Xi). All monsters go straightly to block N.
Now that you know each monster has HP Hi and each tower has a value of attack Di, one attack will cause Di damage (decrease HP by Di). If the HP of a monster is decreased to 0 or below 0, it will die and disappear.
Your task is to calculate the number of monsters surviving from your towers so as to make a plan B.
The path of monsters is a straight line, and there are N blocks on it (numbered from 1 to N continuously). Before enemies come, you have M towers built. Each tower has an attack range [L, R], meaning that it can attack all enemies in every block i, where L<=i<=R. Once a monster steps into block i, every tower whose attack range include block i will attack the monster once and only once. For example, a tower with attack range [1, 3] will attack a monster three times if the monster is alive, one in block 1, another in block 2 and the last in block 3.
A witch helps your enemies and makes every monster has its own place of appearance (the ith monster appears at block Xi). All monsters go straightly to block N.
Now that you know each monster has HP Hi and each tower has a value of attack Di, one attack will cause Di damage (decrease HP by Di). If the HP of a monster is decreased to 0 or below 0, it will die and disappear.
Your task is to calculate the number of monsters surviving from your towers so as to make a plan B.
?
Input
The input contains multiple test cases.
The first line of each case is an integer N (0 < N <= 100000), the number of blocks in the path. The second line is an integer M (0 < M <= 100000), the number of towers you have. The next M lines each contain three numbers, Li, Ri, Di (1 <= Li <= Ri <= N, 0 < Di <= 1000), indicating the attack range [L, R] and the value of attack D of the ith tower. The next line is an integer K (0 < K <= 100000), the number of coming monsters. The following K lines each contain two integers Hi and Xi (0 < Hi <= 10^18, 1 <= Xi <= N) indicating the ith monster’s live point and the number of the block where the ith monster appears.
The input is terminated by N = 0.
The first line of each case is an integer N (0 < N <= 100000), the number of blocks in the path. The second line is an integer M (0 < M <= 100000), the number of towers you have. The next M lines each contain three numbers, Li, Ri, Di (1 <= Li <= Ri <= N, 0 < Di <= 1000), indicating the attack range [L, R] and the value of attack D of the ith tower. The next line is an integer K (0 < K <= 100000), the number of coming monsters. The following K lines each contain two integers Hi and Xi (0 < Hi <= 10^18, 1 <= Xi <= N) indicating the ith monster’s live point and the number of the block where the ith monster appears.
The input is terminated by N = 0.
?
Output
Output one line containing the number of surviving monsters.
?
Sample Input
5
2
1 3 1
5 5 2
5
1 3
3 1
5 2
7 3
9 1
0
?
Sample Output
3
Hint
In the sample, three monsters with origin HP 5, 7 and 9 will survive.
?
Source
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題解及代碼：
官方的思路挺好：
比賽的時候開始使用線段樹來做，結果超時，然后換成樹狀數組來做，使用樹狀數組維護區間和，感覺和官方思路還是差上一線啊。
官方：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#define maxn 100010
using namespace std;
typedef long long ll;
ll delta[maxn];
int main()
{
int n,m,l,r;
ll v;
while(scanf("%d",&n)&&n)
{
memset(delta,0,sizeof(delta[0])*(n+5));
scanf("%d",&m);
for(int i=0;i<m;i++)
{
scanf("%d%d%I64d",&l,&r,&v);
delta[l]+=v;
delta[r+1]-=v;
}
for(int i=2;i<=n;i++)
{
delta[i]+=delta[i-1];
}
for(int i=n-1;i>=1;i--)
{
delta[i]+=delta[i+1];
}
int ans=0;
scanf("%d",&m);
for(int i=0;i<m;i++)
{
scanf("%I64d%d",&v,&r);
if(v>delta[r])
ans++;
}
printf("%d\n",ans);
}
return 0;
}
個人思路：
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
#include <iomanip>
#define INF 99999999
typedef long long LL;
using namespace std;
const int MAX=100000+100;
LL n;
LL c1[MAX],c2[MAX];
LL lowbit(LL x)
{
return x&(-x);
}
void Update(LL x,LL d,LL *c)
{
while(x<=n)
{
c[x]+=d;
x+=lowbit(x);
}
}
LL Query(LL x,LL *c)
{
LL sum=0;
while(x>0)
{
sum+=c[x];
x-=lowbit(x);
}
return sum;
}
int main()
{
LL x,y,d,t,h;
int m;
while(scanf("%I64d",&n)&&n)
{
memset(c1,0,sizeof c1);
memset(c2,0,sizeof c2);
scanf("%d",&m);
for(int i=0; i<m; i++)
{
scanf("%I64d%I64d%I64d",&x,&y,&d);
Update(x,d,c1);
Update(y+1,-d,c1);
Update(x,x*d,c2);
Update(y+1,-(y+1)*d,c2);
}
scanf("%d",&m);
int ans=0;
for(int i=0; i<m; i++)
{
scanf("%I64d%I64d",&h,&x);
t=Query(n,c1)*(n+1)-Query(x-1,c1)*x-Query(n,c2)+Query(x-1,c2);
if(h>t) ans++;
}
printf("%d\n",ans);
}
return 0;
}
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