hdu4970 Killing Monsters （差分数列）

2014多校9 1011

http://acm.hdu.edu.cn/showproblem.php?pid=4970

Killing Monsters

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 331    Accepted Submission(s): 198

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.

 

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.

 

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

2014 Multi-University Training Contest 9

 

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题意：塔防，怪走一条直线，可以分成1~n共n格。给出m个塔的攻击范围（Li~Ri），攻击力Di，怪物走过这格会减少Di血量。给出k个怪物的血量、出生格，求有多少个怪物可以走到终点。

题解：差分数列搞。

粗略一看，是区间加减、区间求和，线段树！会超时，怕了。

再一看，是区间加减完再区间求和，而且求和还是有限制的，就求i~n的和。

用差分数列可以轻松区间加减，差分数列就是b[i]=a[i]-a[i-1]，区间[i,j]加D就是b[i]=b[i]+d  ,  b[j+1]=b[j+1]-d。

但是怎么求和呢？我们把和写出来观察一下：

　　　　　　　　a_n = b_n

　　　　a_n-1 + a_n =2a_n - b_n

a_n-2 + a_n-1 + a_n = 3a_n - 2b_n - b_n-1

看起来很好算的样子！

于是这样就能算（其中an就是a_n ，其中c[i]就是ai加到an的和）：

         ll one=an,many=an;
         for(i=n;i>;i--){
             c[i]=many;
             one-=b[i];
             many+=one;
         }

这样就轻松算啦。

我一开始想到差分数列，不过没仔细想怎么求和，然后就去线段树了，逗乐。后来才发现居然这么好求。

全代码：

 //#pragma comment(linker, "/STACK:102400000,102400000")
 #include<cstdio>
 #include<cmath>
 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<map>
 #include<set>
 #include<stack>
 #include<queue>
 using namespace std;
 #define ll long long
 #define usll unsigned ll
 #define mz(array) memset(array, 0, sizeof(array))
 #define minf(array) memset(array, 0x3f, sizeof(array))
 #define REP(i,n) for(i=0;i<(n);i++)
 #define FOR(i,x,n) for(i=(x);i<=(n);i++)
 #define RD(x) scanf("%d",&x)
 #define RD2(x,y) scanf("%d%d",&x,&y)
 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
 #define WN(x) prllf("%d\n",x);
 #define RE  freopen("D.in","r",stdin)
 #define WE  freopen("1biao.out","w",stdout)
 #define mp make_pair
 #define pb push_back

 const int maxn=;

 int n,m,k;
 int l[maxn],r[maxn],d[maxn];
 int x[maxn];
 ll h[maxn];
 ll b[maxn];
 ll c[maxn];
 ll an=;

 void Update(int L, int R, int x){
     b[L]+=x;
     b[R+]-=x;
     if(R==n)an+=x;
 }

 int main(){
     int i;
     while(scanf("%d",&n)!=EOF){
         if(n==)break;
         scanf("%d",&m);
         mz(b);mz(c);an=;
         REP(i,m) {
             scanf("%d%d%d",&l[i],&r[i],&d[i]);
             Update(l[i],r[i],d[i]);
         }
         ll one=an,many=an;
         for(i=n;i>;i--){
             c[i]=many;
             one-=b[i];
             many+=one;
         }
         //for(i=1;i<=n;i++)printf("%I64d\n",c[i]);
         int ans=;
         //for(i=1;i<=n;i++)printf("(%d,%d),%d\n",i,n,Query(i,n,1,n,1));
         scanf("%d",&k);
         REP(i,k) {
             scanf("%I64d%d",&h[i],&x[i]);
             if(c[x[i]]< h[i])ans++;
         }
         printf("%d\n",ans);
     }
     return ;
 }