2014 Multi-University Training Contest 9#11

2014 Multi-University Training Contest 9#11

Killing MonstersTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 562    Accepted Submission(s): 308

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

521 3 15 5 251 33 15 27 39 10

 

Sample Output

3

Hint

In the sample, three monsters with origin HP 5, 7 and 9 will survive.

 

简单的线段树，区间更新就可以！

 

  1 #include<cstdio>
  2 
  3 using namespace std;
  4 
  5 #define maxn 100000+10
  6 
  7 typedef long long LL;
  8 
  9 struct node{
 10 
 11     int l,r,m;
 12 
 13     LL sum,mark;
 14 
 15 }T[maxn<<2];
 16 
 17 int a[maxn];
 18 
 19 void build(int id,int l,int r){
 20 
 21      T[id].l=l;   T[id].r=r;  T[id].m=(l+r)>>1;  T[id].mark=0;
 22 
 23      if(l==r)   { T[id].sum=a[l]; return;  }
 24 
 25      int m=(l+r)>>1;
 26 
 27      build(id<<1,l,m);  build((id<<1)+1,m+1,r);
 28 
 29      T[id].sum=(T[id<<1].sum+T[(id<<1)+1].sum);
 30 
 31 }
 32 
 33 void update(int id,int l,int r,int val){
 34 
 35      if(T[id].l==l&&T[id].r==r){
 36 
 37         T[id].mark+=val; return ;
 38 
 39      }
 40 
 41      T[id].sum+=(LL)val*(r-l+1);
 42 
 43      if(T[id].m>=r)
 44 
 45           update(id<<1,l,r,val);
 46 
 47      else if(T[id].m<l)
 48 
 49           update((id<<1)+1,l,r,val);
 50 
 51      else{
 52 
 53           update(id<<1,l,T[id].m,val);
 54 
 55           update((id<<1)+1,T[id].m+1,r,val);
 56 
 57      }
 58 
 59 }
 60 
 61 LL query(int id,int l,int r){
 62 
 63     if(T[id].l==l&&T[id].r==r)  return T[id].sum+T[id].mark*(LL)(r-l+1);
 64 
 65     if(T[id].mark!=0) {
 66 
 67         T[id<<1].mark+=T[id].mark;
 68 
 69         T[(id<<1)+1].mark+=T[id].mark;
 70 
 71         T[id].sum+=(LL)(T[id].r-T[id].l+1)*T[id].mark;  T[id].mark=0;
 72 
 73     }
 74 
 75 
 76 
 77     if(T[id].m>=r){
 78 
 79           return query(id<<1,l,r);
 80 
 81     }
 82 
 83     else if(T[id].m<l){
 84 
 85           return query((id<<1)+1,l,r);
 86 
 87     }
 88 
 89     else{
 90 
 91           return query(id<<1,l,T[id].m)+query((id<<1)+1,T[id].m+1,r);
 92 
 93     }
 94 
 95 }
 96 
 97 int main(){
 98 
 99     int n,Q,m;  char str[8];   int b,c,d,l,r,v;
100 
101     while(scanf("%d",&n)==1&&n){
102 
103             scanf("%d",&Q);
104 
105           for(int i=1;i<=n;i++){
106 
107              a[i] = 0;
108 
109           }
110 
111           build(1,1,n);
112 
113           for(int i=0;i<Q;i++){
114 
115                scanf("%d%d%d",&l,&r,&v);
116 
117                     update(1,l,r,v);
118 
119                }
120 
121           scanf("%d",&m);
122 
123           LL h;
124 
125           int x;
126 
127           int ans = 0;
128 
129           for(int i = 0;i<m;i++)
130 
131           {
132 
133               scanf("%I64d%d",&h,&x);
134 
135               if(query(1,x,n)<h) ans++;
136 
137           }
138 
139           printf("%d\n",ans);
140 
141     }
142 
143     return 0;
144 
145 }

 

转载于:https://www.cnblogs.com/haohaooo/p/4035334.html