Under Control
Time Limit: 2000ms
Memory Limit: 65536KB
This problem will be judged on ZJU. Original ID: 2605
64-bit integer IO format: %lld Java class name: Main
In a game of Civilization III the area controlled by a city is defined by its culture level. The game proceeds on a rectangular grid. A city occupies one grid square. Each city has aculture level which is a non-negative integer number.
A city with a culture level 0 controls its own square and eight adjacent squares. A city with a culture level 1 additionally controls all squares that share a side with those squares (a total of 9 + 12 = 21 squares). Generally, if a city with a culture level i controls the set A of squares, a city with the same location and a culture level i + 1 would control all these squares and also squares that share a side with at least one square from A.
The picture on the left shows the sets of squares controlled by cities with culture levels of 0, 1 and 2.
The area controlled by the civilization is defined as follows. Consider the total area controlled by all its cities. The civilization area is the smallest set of squares, such that it contains all the squares controlled by some city, and its complement contains no hanging squares. A square x of a set B is called hanging if there is no 2 * 2 square in B that contains square x.
Calculate the total area controlled by a civilization, given the locations of all its cities on a map. You may consider that the map is infinite and that there are no other civilizations.
Input
The input consists of several test cases. In each case, the first line contains an integral number n - the number of the cities of a civilization (1 <= n <= 50). Next n lines describe cities. Each city is described with its integer coordinates (xi, yi) and its culture level ci. Coordinates do not exceed 109 by their absolute value, culture level does not exceed 10. The input ends up with a case where n = 0. Do not proceed this case.
Output
Output the total number of squares controlled by a civilization, each case in a single line.
Sample Input
2
0 0 1
4 -3 0
0
31
Source
Andrew Stankevich's Contest #7
Author
Andrew Stankevich
解題：亂搞，主要是坐标的範圍太大，是以用set存已經被占領了的點。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <cmath>
5 #include <algorithm>
6 #include <climits>
7 #include <vector>
8 #include <queue>
9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define INF 0x3f3f3f3f
15 #define pii pair<int,int>
16 using namespace std;
17 const int maxn = 100000;
18 const int dir[4][2] = {-1,0,0,1,0,-1,1,0};
19 pii p[maxn];
20 set< pii > s;
21 int tot;
22 void solve(int x,int y,int v){
23 int a = 3 + v*2,k = a>>1;
24 int tx = x - 1;
25 int ty = y + k;
26 for(int i = 0; i < k; ++i){
27 for(int j = 0; j < 3+2*i; ++j){
28 pii tmp = make_pair(tx+j,ty);
29 if(s.count(tmp) == 1) continue;
30 s.insert(tmp);
31 p[tot++] = tmp;
32 }
33 --tx;
34 --ty;
35 }
36 tx = x - k;
37 ty = y;
38 for(int i = 0; i < a; ++i){
39 pii tmp = make_pair(tx+i,ty);
40 if(s.count(tmp) == 1) continue;
41 s.insert(tmp);
42 p[tot++] = tmp;
43 }
44 tx = x - k;
45 ty = y - 1;
46 for(int i = 0; i < k; ++i){
47 for(int j = 0; j < a; ++j){
48 pii tmp = make_pair(tx+j,ty);
49 if(s.count(tmp) == 1) continue;
50 s.insert(tmp);
51 p[tot++] = tmp;
52 }
53 a -= 2;
54 tx++;
55 ty--;
56 }
57 }
58 bool iswhite(int x,int y){
59 pii tmp = make_pair(x,y);
60 return s.count(tmp) == 0;
61 }
62 void go(int x,int y){
63 bool p1 = iswhite(x-1,y+1);
64 bool p2 = iswhite(x,y+1);
65 bool p3 = iswhite(x+1,y+1);
66 bool p4 = iswhite(x-1,y);
67 bool p5 = iswhite(x+1,y);
68 bool p6 = iswhite(x-1,y-1);
69 bool p7 = iswhite(x,y-1);
70 bool p8 = iswhite(x+1,y-1);
71
72 if(p1 && p2 && p4) return;
73 if(p2 && p3 && p5) return;
74 if(p4 && p6 && p7) return;
75 if(p5 && p7 && p8) return;
76
77 pii tmp = make_pair(x,y);
78 p[tot++] = tmp;
79 s.insert(tmp);
80 }
81 int main(){
82 int n,x,y,v;
83 while(scanf("%d",&n),n){
84 s.clear();
85 for(int i = tot = 0; i < n; ++i){
86 scanf("%d %d %d",&x,&y,&v);
87 solve(x,y,v);
88 }
89 for(int i = 0; i < tot; ++i){
90 for(int j = 0; j < 4; ++j){
91 int tx = p[i].first + dir[j][0];
92 int ty = p[i].second + dir[j][1];
93 if(iswhite(tx,ty)) go(tx,ty);
94 }
95 }
96 printf("%d\n",s.size());
97 }
98 return 0;
99 } View Code
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