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UVa 1641 ASCII Area (计算几何,水题)
题意:给定一个矩阵,里面有一个多边形,求多边形的面积。
析:因为是在格子里,并且这个多边形是很规则的,所以所有格子不是全属于多边形就是全不属于,或者一半,并且我们可以根据"/"和“\”的数目来知道,如果是奇数,那么就是属于,
偶数就是不属于。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <string>#include <cstdlib>#include <cmath>#include <iostream>#include <cstring>#include <set>#include <queue>#include <algorithm>#include <vector>#include <map>#include <cctype>#include <cmath>#include <stack>#define frer freopen("in.txt", "r", stdin)#define frew freopen("out.txt", "w", stdout)using namespace std;typedef long long LL;typedef pair<int, int> P;const int INF = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f;const double PI = acos(-1.0);const double eps = 1e-8;const int maxn = 1e2 + 5;const int mod = 1e9 + 7;const char *mark = "+-*";const int dr[] = {-1, 0, 1, 0};const int dc[] = {0, 1, 0, -1};const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};int n, m;const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};inline int Min(int a, int b){ return a < b ? a : b; }inline int Max(int a, int b){ return a > b ? a : b; }inline LL Min(LL a, LL b){ return a < b ? a : b; }inline LL Max(LL a, LL b){ return a > b ? a : b; }inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m;}char s[maxn];int main(){ while(scanf("%d %d", &n, &m) == 2){ int ans = 0; for(int i = 0; i < n; ++i){ int cnt = 0; scanf("%s", s); for(int j = 0; j < m; ++j){ if(s[j] == ‘/‘ || s[j] == ‘\\‘) ++ans, ++cnt; else if(cnt & 1) ans += 2; } } printf("%d\n", ans/2); } return 0;}
UVa 1641 ASCII Area (计算几何,水题)
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