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题意:给个棋盘,你可以在棋盘的边缘处放2个蓝色棋子2个黄色棋子,问连接2组同色棋子的最小代价,如果线路交叉,输-1。
析:交叉么,可以把它们看成是两条线段,然后如果相交就是不行的,但是有几种特殊情况,比如都在同一行或同一列,要特殊考虑这种情况。
1122,1212,2211,2121,1221,2112.这是几种特殊的,然后其他的就可以用判交叉来算了,然后最短路就是横纵坐标相减的绝对值加2.
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 <stack> #define freopenr freopen("in.txt", "r", stdin) #define freopenw 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 LL LNF = 100000000000000000; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 3e5 + 5; const int mod = 1e9 + 7; const char *mark = "+-*"; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; int n, m; inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } inline LL Max(LL a, LL b){ return a < b ? b : a; } inline LL Min(LL a, LL b){ return a > b ? b : a; } inline int Max(int a, int b){ return a < b ? b : a; } inline int Min(int a, int b){ return a > b ? b : a; } char s[15][15]; vector<P> v1; vector<P> v2; inline int mult(const P &a, const P &b, const P &c){ return (a.first-c.first)*(b.second-c.second) - (b.first-c.first)*(a.second-c.second); } bool intersection(const P &aa, const P &bb, const P &cc, const P &dd){ if(Max(aa.first, bb.first) < Min(cc.first, dd.first)) return false; if(Max(aa.second, bb.second) < Min(cc.second, dd.second)) return false; if(Max(cc.first, dd.first) < Min(aa.first, bb.first)) return false; if(Max(cc.second, dd.second) < Min(aa.second, bb.second)) return false; if(mult(cc, bb, aa) * mult(bb, dd, aa) < 0) return false; if(mult(aa, dd, cc) * mult(dd, bb, cc) < 0) return false; return true; } bool solve(int a, int b, int c, int d){ if(a < b && b < c && c < d){ printf("%d\n", b-a+d-c+2); return true; } else if(c < d && d < a && a < b){ printf("%d\n", b-a+d-c+2); return true; } else if(a < c && c < d && d < b){ printf("%d\n", b-a+d-c+4); return true; } else if(c < a && a < b && b < d){ printf("%d\n", b-a+d-c+4); return true; } else if(a < c && c < b && b < d){ printf("-1\n"); return true; } else if(c < a && a < d && d < b){ printf("-1\n"); return true; } return false; } bool judge1(){ bool ok = true; if(v1[0].first == v1[1].first && v1[1].first == v2[0].first && v1[1].first == v2[1].first){ if(solve(v1[0].second, v1[1].second, v2[0].second, v2[1].second)) return true; } else if(v1[0].second == v1[1].second && v1[1].second == v2[0].second && v1[1].second == v2[1].second){ if(solve(v1[0].first, v1[1].first, v2[0].first, v2[1].first)) return true; } return false; } int main(){ int T; cin >> T; while(T--){ v1.clear(); v2.clear(); scanf("%d", &n); for(int i = 1; i <= n; ++i) scanf("%s", s[i]+1); for(int i = 1; i <= n; ++i) for(int j = 1; j <= n; ++j) if(s[i][j] == ‘1‘) v1.push_back(P(i, j)); else if(s[i][j] == ‘2‘) v2.push_back(P(i, j)); sort(v1.begin(), v1.end()); sort(v2.begin(), v2.end()); if(judge1()) continue; if(intersection(v1[0], v1[1], v2[0], v2[1])){ printf("-1\n"); continue; } printf("%d\n", abs(v1[1].first-v1[0].first)+abs(v1[1].second-v1[0].second)+abs(v2[1].first-v2[0].first)+abs(v2[1].second-v2[0].second)+2); } return 0; }
UVaLive 6693 Flow Game (计算几何,线段相交)
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原文地址:http://www.cnblogs.com/dwtfukgv/p/5801381.html