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UVaLive 6693 Flow Game (计算几何,线段相交)

时间:2016-08-24 06:30:34      阅读:158      评论:0      收藏:0      [点我收藏+]

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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

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