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 #include #include #include #include #include #include #include #include #include #include #include #include #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef pair 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 = 0 && c b ? b : a; } inline int Max(int a, int b){ return a b ? b : a; } char s[15][15]; vector v1; vector 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) > 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; }
v1; vector
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) > 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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