2014多校9 1011
http://acm.hdu.edu.cn/showproblem.php?pid=4970
Killing MonstersTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) Problem Description Kingdom Rush is a popular TD game, in which you should build some The path of monsters is a straight line, and there A witch helps your enemies and makes Now that Input The input contains multiple test cases. The first line of each case is an integer N (0 < N <= 100000), The input is terminated by N = 0. Output Output one line containing the number of surviving monsters. Sample Input 5 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 Recommend hujie | We have carefully selected several similar problems for you: 4970 4969 4968 4967 4966 |
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：
an = bn
an-1 + an =2an - bn
an-2 + an-1 + an = 3an - 2bn - bn-1
看起来很好算的样子！
于是这样就能算（其中an就是an ，其中c[i]就是ai加到an的和）：
1 ll one=an,many=an;
2 for(i=n;i>0;i--){
3 c[i]=many;
4 one-=b[i];
5 many+=one;
6 }
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：
1 //#pragma comment(linker, "/STACK:102400000,102400000")
2 #include<cstdio>
3 #include<cmath>
4 #include<iostream>
5 #include<cstring>
6 #include<algorithm>
7 #include<cmath>
8 #include<map>
9 #include<set>
10 #include<stack>
11 #include<queue>
12 using namespace std;
13 #define ll long long
14 #define usll unsigned ll
15 #define mz(array) memset(array, 0, sizeof(array))
16 #define minf(array) memset(array, 0x3f, sizeof(array))
17 #define REP(i,n) for(i=0;i<(n);i++)
18 #define FOR(i,x,n) for(i=(x);i<=(n);i++)
19 #define RD(x) scanf("%d",&x)
20 #define RD2(x,y) scanf("%d%d",&x,&y)
21 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
22 #define WN(x) prllf("%d\n",x);
23 #define RE freopen("D.in","r",stdin)
24 #define WE freopen("1biao.out","w",stdout)
25 #define mp make_pair
26 #define pb push_back
27
28 const int maxn=111111;
29
30 int n,m,k;
31 int l[maxn],r[maxn],d[maxn];
32 int x[maxn];
33 ll h[maxn];
34 ll b[maxn];
35 ll c[maxn];
36 ll an=0;
37
38 void Update(int L, int R, int x){
39 b[L]+=x;
40 b[R+1]-=x;
41 if(R==n)an+=x;
42 }
43
44 int main(){
45 int i;
46 while(scanf("%d",&n)!=EOF){
47 if(n==0)break;
48 scanf("%d",&m);
49 mz(b);mz(c);an=0;
50 REP(i,m) {
51 scanf("%d%d%d",&l[i],&r[i],&d[i]);
52 Update(l[i],r[i],d[i]);
53 }
54 ll one=an,many=an;
55 for(i=n;i>0;i--){
56 c[i]=many;
57 one-=b[i];
58 many+=one;
59 }
60 //for(i=1;i<=n;i++)printf("%I64d\n",c[i]);
61 int ans=0;
62 //for(i=1;i<=n;i++)printf("(%d,%d),%d\n",i,n,Query(i,n,1,n,1));
63 scanf("%d",&k);
64 REP(i,k) {
65 scanf("%I64d%d",&h[i],&x[i]);
66 if(c[x[i]]< h[i])ans++;
67 }
68 printf("%d\n",ans);
69 }
70 return 0;
71 }
hdu4970 Killing Monsters （差分数列）,布布扣,bubuko.com