SRM 446(1-250pt, 1-500pt)

嗯。。。。今天的500确实比较好

DIV1 250

　　模拟。。。略

  1 // BEGIN CUT HERE
  2 /*
  3  * Author:  plum rain
  4  * score :
  5  */
  6 /*
  7 
  8  */
  9 // END CUT HERE
 10 #line 11 "CubeWalking.cpp"
 11 #include <sstream>
 12 #include <stdexcept>
 13 #include <functional>
 14 #include <iomanip>
 15 #include <numeric>
 16 #include <fstream>
 17 #include <cctype>
 18 #include <iostream>
 19 #include <cstdio>
 20 #include <vector>
 21 #include <cstring>
 22 #include <cmath>
 23 #include <algorithm>
 24 #include <cstdlib>
 25 #include <set>
 26 #include <queue>
 27 #include <bitset>
 28 #include <list>
 29 #include <string>
 30 #include <utility>
 31 #include <map>
 32 #include <ctime>
 33 #include <stack>
 34 
 35 using namespace std;
 36 
 37 #define clr0(x) memset(x, 0, sizeof(x))
 38 #define clr1(x) memset(x, -1, sizeof(x))
 39 #define pb push_back
 40 #define sz(v) ((int)(v).size())
 41 #define all(t) t.begin(),t.end()
 42 #define zero(x) (((x)>0?(x):-(x))<eps)
 43 #define out(x) cout<<#x<<":"<<(x)<<endl
 44 #define tst(a) cout<<a<<" "
 45 #define tst1(a) cout<<#a<<endl
 46 #define CINBEQUICKER std::ios::sync_with_stdio(false)
 47 
 48 typedef vector<int> vi;
 49 typedef vector<string> vs;
 50 typedef vector<double> vd;
 51 typedef pair<int, int> pii;
 52 typedef long long int64;
 53 
 54 const double eps = 1e-8;
 55 const double PI = atan(1.0)*4;
 56 const int inf = 2139062143 / 2;
 57 int temp[4][2] = {{0, -1}, {1, 0}, {0, 1}, {-1, 0}};
 58 
 59 class CubeWalking
 60 {
 61     public:
 62         string finalPosition(string mov){
 63             int n = sz(mov);
 64             int x = 1, y = 1, dic = 0;
 65             for (int i = 0; i < n; ++ i){
 66                 if (mov[i] == 'L') dic = (dic + 3) % 4;
 67                 else if (mov[i] == 'R') dic = (dic + 1) % 4;
 68                 if (mov[i] == 'W') x = (x + temp[dic][0] + 3) % 3, y = (y + temp[dic][1] + 3) % 3;
 69             }
 70             if (x == 1 && y == 1) return "GREEN";
 71             if (!x && y == 1) return "BLUE";
 72             if (!y && x == 1) return "BLUE";
 73             if (y == 2 && x == 1) return "BLUE";
 74             if (x == 2 && y == 1) return "BLUE";
 75             return "RED";
 76         }
 77         
 78 // BEGIN CUT HERE
 79     public:
 80     void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); }
 81     private:
 82     template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }
 83     void verify_case(int Case, const string &Expected, const string &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
 84     void test_case_0() { string Arg0 = "LLRR"; string Arg1 = "GREEN"; verify_case(0, Arg1, finalPosition(Arg0)); }
 85     void test_case_1() { string Arg0 = "WWWWWWWWWWWW"; string Arg1 = "GREEN"; verify_case(1, Arg1, finalPosition(Arg0)); }
 86     void test_case_2() { string Arg0 = "WLWRW"; string Arg1 = "RED"; verify_case(2, Arg1, finalPosition(Arg0)); }
 87     void test_case_3() { string Arg0 = "WWLLWRWLWLLRWW"; string Arg1 = "BLUE"; verify_case(3, Arg1, finalPosition(Arg0)); }
 88 
 89 // END CUT HERE
 90 
 91 };
 92 
 93 // BEGIN CUT HERE
 94 int main()
 95 {
 96 //    freopen( "a.out" , "w" , stdout );    
 97     CubeWalking ___test;
 98     ___test.run_test(-1);
 99        return 0;
100 }
101 // END CUT HERE
View Code

 

DIV1 500

题意：给定一个有n个点的带权图，有一个蚂蚁要从点0到点1，它每秒走stp步，最多能走tim秒，走的时间可以少于tim秒，但一定要是整数秒。问，若蚂蚁得分为它通过的边权之和(通过某边两次则得两次分)，则求最大得分，若不能用整数秒走到1并停下，输出IMPOSSIBLE。

　　　n <= 50, stp <= 100, tim <= 10^9。

解法：首先，如果要求一定要走tim，那就很简单了，用矩阵乘法快速幂可以解决。

　　　至于走的时间可以少于tim，处理方法就是加一条1到1，长度为1s，边权为0的自环。

　　　然后，首先用矩阵乘法快速幂算出用时1s，从i点走到j点得分最高为多少。再赋值A[1][1] = max(A[1][1], 0)(若A[1][1]大于0则不用赋值)，然后再调用矩阵乘法快速幂即可。

tag:graph, matrix, 快速幂, good

  1 // BEGIN CUT HERE
  2 /*
  3  * Author:  plum rain
  4  * score :
  5  */
  6 /*
  7 
  8  */
  9 // END CUT HERE
 10 #line 11 "AntOnGraph.cpp"
 11 #include <sstream>
 12 #include <stdexcept>
 13 #include <functional>
 14 #include <iomanip>
 15 #include <numeric>
 16 #include <fstream>
 17 #include <cctype>
 18 #include <iostream>
 19 #include <cstdio>
 20 #include <vector>
 21 #include <cstring>
 22 #include <cmath>
 23 #include <algorithm>
 24 #include <cstdlib>
 25 #include <set>
 26 #include <queue>
 27 #include <bitset>
 28 #include <list>
 29 #include <string>
 30 #include <utility>
 31 #include <map>
 32 #include <ctime>
 33 #include <stack>
 34 
 35 using namespace std;
 36 
 37 #define clr0(x) memset(x, 0, sizeof(x))
 38 #define clr1(x) memset(x, -1, sizeof(x))
 39 #define pb push_back
 40 #define sz(v) ((int)(v).size())
 41 #define all(t) t.begin(),t.end()
 42 #define zero(x) (((x)>0?(x):-(x))<eps)
 43 #define out(x) cout<<#x<<":"<<(x)<<endl
 44 #define tst(a) cout<<a<<" "
 45 #define tst1(a) cout<<#a<<endl
 46 #define CINBEQUICKER std::ios::sync_with_stdio(false)
 47 
 48 typedef long long int64;
 49 typedef vector<int> vi;
 50 typedef vector<string> vs;
 51 typedef vector<double> vd;
 52 typedef pair<int, int> pii;
 53 typedef int64 mtx[500][500];
 54 
 55 const double eps = 1e-8;
 56 const double PI = atan(1.0)*4;
 57 const int inf = 2139062143 / 2;
 58 const int64 temp = 0 - (1LL<<62);
 59 
 60 class AntOnGraph
 61 {
 62     public:
 63         int dit[1000], n;
 64         void mtx_eql(mtx &a, mtx b)
 65         {
 66             for (int i = 0; i < n; ++ i) for (int j = 0; j < n; ++ j) a[i][j] = b[i][j];
 67         }
 68         void mtx_mul(mtx &a, mtx b)
 69         {
 70             mtx ret;
 71             for (int i = 0; i < n; ++ i)
 72                 for (int j = 0; j < n; ++ j){
 73                     ret[i][j] = temp;
 74                     for (int k = 0; k < n; ++ k) 
 75                         if (a[i][k] != temp && b[k][j] != temp)
 76                             ret[i][j] = max(ret[i][j], a[i][k] + b[k][j]);
 77                 }
 78 
 79             mtx_eql(a, ret);
 80         }
 81         void mtx_pow(mtx &an, int64 num)
 82         {
 83             if (num == 1) return;
 84 
 85             mtx ret; mtx_eql(ret, an);
 86             -- num;
 87 
 88             while (num){
 89                 if (num & 1) mtx_mul(ret, an);
 90                 num >>= 1;
 91                 mtx_mul(an, an);
 92             }
 93             
 94             mtx_eql(an, ret);
 95         }
 96 
 97         string gao(int64 x)
 98         {
 99             int len = 0;
100             bool u = 0;
101             if (x < 0) x = 0 - x, u = 1;
102 
103             while (x){
104                 dit[len++] = x % 10;
105                 x /= 10;
106             }
107 
108             if (!len) return "0";
109             string ret; ret.clear();
110             if (u) ret.pb ('-');
111             for (int i = len-1; i >= 0; -- i) ret.pb (dit[i] + '0');
112             return ret;
113         }
114         string maximumBonus(vector <string> p0, vector <string> p1, vector <string> p2, int stp, int tim){
115             n = sz(p1);
116             int cnt = 0;
117             mtx an;
118             for (int i = 0; i < n; ++ i)
119                 for (int j = 0; j < n; ++ j){
120                     cnt = 100 * (p0[i][j]-'0') + 10 * (p1[i][j] - '0') + p2[i][j] - '0';
121                     if (cnt == 0) an[i][j] = temp;
122                     else an[i][j] = cnt - 500;
123                 }
124 
125             mtx_pow(an, (int64)stp);
126             if (an[1][1] < 0) an[1][1] = 0;
127 
128             mtx_pow(an, (int64)tim);
129             if (an[0][1] == temp) return "IMPOSSIBLE";
130             else return gao(an[0][1]);
131         }
132         
133 // BEGIN CUT HERE
134     public:
135     void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); if ((Case == -1) || (Case == 4)) test_case_4(); }
136     private:
137     template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }
138     void verify_case(int Case, const string &Expected, const string &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
139     void test_case_0() { string Arr0[] = {"05","50"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); string Arr1[] = {"00","00"}; vector <string> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); string Arr2[] = {"01","10"}; vector <string> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[0]))); int Arg3 = 3; int Arg4 = 2; string Arg5 = "3"; verify_case(0, Arg5, maximumBonus(Arg0, Arg1, Arg2, Arg3, Arg4)); }
140     void test_case_1() { string Arr0[] = {"05","50"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); string Arr1[] = {"00","00"}; vector <string> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); string Arr2[] = {"01","10"}; vector <string> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[0]))); int Arg3 = 2; int Arg4 = 3; string Arg5 = "IMPOSSIBLE"; verify_case(1, Arg5, maximumBonus(Arg0, Arg1, Arg2, Arg3, Arg4)); }
141     void test_case_2() { string Arr0[] = {"0550","0000","0005","5000"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); string Arr1[] = {"0000","0000","0000","0000"}; vector <string> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); string Arr2[] = {"0110","0000","0001","1000"}; vector <string> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[0]))); int Arg3 = 7; int Arg4 = 9; string Arg5 = "49"; verify_case(2, Arg5, maximumBonus(Arg0, Arg1, Arg2, Arg3, Arg4)); }
142     void test_case_3() { string Arr0[] = {"0540","0000","0004","4000"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); string Arr1[] = {"0090","0000","0009","9000"}; vector <string> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); string Arr2[] = {"0190","0000","0009","9000"}; vector <string> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[0]))); int Arg3 = 7; int Arg4 = 9; string Arg5 = "-5"; verify_case(3, Arg5, maximumBonus(Arg0, Arg1, Arg2, Arg3, Arg4)); }
143     void test_case_4() { string Arr0[] = {"079269665406","506042219642","720809987956",
144  "315099331918","952306192584","406390344278",
145  "999241035142","370785209189","728363760165",
146  "019367419000","279718007804","610188263490"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); string Arr1[] = {"038604914953","804585763146","350629473403",
147  "028096403898","575205051686","427800322647",
148  "655168017864","582553303278","980402170192",
149  "620737714031","686142310509","092421645460"}; vector <string> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); string Arr2[] = {"063231394554","109852259379","740182746422",
150  "853015982521","476805512496","898530717993",
151  "430534005863","440162807186","132879980431",
152  "685312177072","780267345400","959947501200"}; vector <string> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[0]))); int Arg3 = 37; int Arg4 = 1221; string Arg5 = "20992235"; verify_case(4, Arg5, maximumBonus(Arg0, Arg1, Arg2, Arg3, Arg4)); }
153 
154 // END CUT HERE
155 
156 };
157 
158 // BEGIN CUT HERE
159 int main()
160 {
161 //    freopen( "a.out" , "w" , stdout );    
162     AntOnGraph ___test;
163     ___test.run_test(-1);
164        return 0;
165 }
166 // END CUT HERE
View Code

 

 

 

posted @ 2014-02-16 17:22  Plumrain  阅读(270)  评论(0)  编辑  收藏  举报