SRM 149
DIV2 1000pt
题意：
对于n个人，第i人有pi的钱。将他们分成不超过四个组，每组统一交费x，对每个人，若他拥有的钱超过x则交费，否则不交费。问最多能使这些人交多少钱。
1<= n <= 50，0 <= pi <= 1000。
tag:greedy，think
解法：枚举所有分组情况，每组交费x为该组中最小的pi。如果分组不足四组，补足四组，每组均含一个人钱数为0的人。
Ps：感觉自己的代码写得比题解舒服.....
1 #line 7 "Pricing.cpp" 2 #include <cstdlib> 3 #include <cctype> 4 #include <cstring> 5 #include <cstdio> 6 #include <cmath> 7 #include <algorithm> 8 #include <vector> 9 #include <iostream> 10 #include <sstream> 11 #include <set> 12 #include <queue> 13 #include <fstream> 14 #include <numeric> 15 #include <iomanip> 16 #include <bitset> 17 #include <list> 18 #include <stdexcept> 19 #include <functional> 20 #include <string> 21 #include <utility> 22 #include <map> 23 #include <ctime> 24 #include <stack> 25 26 using namespace std; 27 28 #define CLR(x) memset(x, 0, sizeof(x)) 29 #define PB push_back 30 #define SZ(v) ((int)(v).size()) 31 #define out(x) cout<<#x<<":"<<(x)<<endl 32 #define tst(a) cout<<#a<<endl 33 #define CINBEQUICKER std::ios::sync_with_stdio(false) 34 35 typedef vector<int> VI; 36 typedef vector<string> VS; 37 typedef vector<double> VD; 38 typedef long long int64; 39 40 const double eps = 1e-8; 41 const double PI = atan(1.0)*4; 42 const int maxint = 2139062143; 43 44 inline int MyMod( int a , int b ) { return (a%b+b)%b;} 45 46 int n, ans; 47 48 bool cmp(int x, int y) 49 { 50 return x > y; 51 } 52 53 class Pricing 54 { 55 public: 56 int maxSales(vector <int> p){ 57 p.PB(0), p.PB(0), p.PB(0); 58 n = SZ (p); 59 if (!n) return 0; 60 ans = 0; 61 sort (p.begin(), p.end(), cmp); 62 63 for (int i = 0; i < n; ++ i) 64 for (int j = i+1; j < n; ++ j) 65 for (int k = j+1; k < n; ++ k) 66 for (int a = k+1; a < n; ++ a){ 67 int tmp = 0; 68 tmp = p[i] * (i+1) + p[j] * (j-i); 69 tmp += p[k] * (k-j) + p[a] * (a-k); 70 ans = max (ans, tmp); 71 } 72 73 return (ans); 74 } 75 //by plum rain 76 };
DIV1 500pt
题意：
字符串匹配。给出字典和一个字符串，询问该字符串是否能用字典里的单词表示，有一种表示方法还是多种表示方法，如果只有一种，输出表示方法。
tag:dp
Ps:以后写dp一定要提前写好状态转移方程- -，错了好多次- -。还有，此题要用long long。
1 #line 7 "MessageMess.cpp" 2 #include <cstdlib> 3 #include <cctype> 4 #include <cstring> 5 #include <cstdio> 6 #include <cmath> 7 #include <algorithm> 8 #include <vector> 9 #include <iostream> 10 #include <sstream> 11 #include <set> 12 #include <queue> 13 #include <fstream> 14 #include <numeric> 15 #include <iomanip> 16 #include <bitset> 17 #include <list> 18 #include <stdexcept> 19 #include <functional> 20 #include <string> 21 #include <utility> 22 #include <map> 23 #include <ctime> 24 #include <stack> 25 26 using namespace std; 27 28 #define CLR(x) memset(x, 0, sizeof(x)) 29 #define PB push_back 30 #define SZ(v) ((int)(v).size()) 31 #define out(x) cout<<#x<<":"<<(x)<<endl 32 #define tst(a) cout<<#a<<endl 33 #define CINBEQUICKER std::ios::sync_with_stdio(false) 34 35 typedef vector<int> VI; 36 typedef vector<string> VS; 37 typedef vector<double> VD; 38 typedef long long int64; 39 40 const double eps = 1e-8; 41 const double PI = atan(1.0)*4; 42 const int maxint = 2139062143; 43 const int N = 55; 44 45 inline int MyMod( int a , int b ) { return (a%b+b)%b;} 46 47 int64 dp[N], pat[N]; 48 49 class MessageMess 50 { 51 public: 52 string restore(vector <string> d, string s){ 53 CLR (dp), CLR (pat); 54 int m = SZ (d), n = SZ (s); 55 for (int i = 0; i < n; ++ i){ 56 for (int j = 0; j < m; ++ j){ 57 int size = SZ (d[j]); 58 int pos = i - size; 59 if (pos < -1) continue; 60 string tmp(s, pos+1, size); 61 if (tmp == d[j]){ 62 if (pos == -1) 63 dp[i] += 1, pat[i] = j; 64 else if (dp[pos] > 0) 65 dp[i] += dp[pos], pat[i] = j; 66 } 67 } 68 } 69 70 if (!dp[n-1]) return "IMPOSSIBLE!"; 71 if (dp[n-1] != 1) return "AMBIGUOUS!"; 72 VS ans; ans.clear(); 73 int pos = n - 1; 74 while (pos >= 0){ 75 int num = pat[pos]; 76 ans.PB(d[num]); 77 pos -= SZ (d[num]); 78 } 79 int len = SZ (ans); 80 string ret; ret.clear(); 81 ret += ans[len-1]; 82 for (int i = len-2; i >= 0; -- i) 83 ret += " " + ans[i]; 84 return (ret); 85 } 86 87 //by plum rain 88 };
DIV1 1000pt
题意：
在平面坐标系XOY中，以xi严格递增的顺序给出一些点的坐标(xi, yi)，然后将点(xi, yi)与(x[i+1], y[i+1])连接，得到一条折线。给定整数p，求在这条折线上的某一线段，起点终点分别为(a, b), (a+p, b')，使这一线段平均y值最大，输出最大的平均y值。
x, y范围[0, 10000]。
tag:think, (三分法), good
解法：
没做出来。官方题解提供了两种方法，第一种方法似乎要用三分法，还不太懂- -。
第二种方法是，考虑线段(a, b)——(a+p, b')在折线上移动，记该线段的平均y值为tmp。该线段从原位置移动到(a+1, c)和(a+p+1, c')，则tmp' = tmp - (f(a+1) + f(a)) / 2 + (f(a+p+1) + f(a+p)) / 2), ans = max (tmp, ans)(其中，f(a)表示折线段的在x = a时的y值，ans表示答案)。
然后考虑到线段(a, b)——(a+1, c)和线段(a+p, b')——(a+p+1, c')的斜率可能不同，所以答案线段也有可能出现在移动途中。记两线段y值相等的点为(x1, y0) 和 (x2, y0), 则ans = max (ans, tmp' + ((y0 + c) / 2)* (a+1 - x1) - ((y0 + c') / 2) * (a+p+1 - x2))。
PS：
返回double类型的要注意精度，并且，如果你使用topcoder的插件，在你自己的电脑上判断答案对错时，有可能因为精度问题而将你的正确答案判为错，输出一下返回值看一下是否正确就行。
1 /* 2 * Author: plum rain 3 */ 4 #line 11 "topcoder.cpp" 5 #include <cstdlib> 6 #include <cctype> 7 #include <cstring> 8 #include <cstdio> 9 #include <cmath> 10 #include <algorithm> 11 #include <vector> 12 #include <iostream> 13 #include <sstream> 14 #include <set> 15 #include <queue> 16 #include <fstream> 17 #include <numeric> 18 #include <iomanip> 19 #include <bitset> 20 #include <list> 21 #include <stdexcept> 22 #include <functional> 23 #include <string> 24 #include <utility> 25 #include <map> 26 #include <ctime> 27 #include <stack> 28 29 using namespace std; 30 31 #define CLR(x) memset(x, 0, sizeof(x)) 32 #define PB push_back 33 #define SZ(v) ((int)(v).size()) 34 #define out(x) cout<<#x<<":"<<(x)<<endl 35 #define tst(a) cout<<#a<<endl 36 #define CINBEQUICKER std::ios::sync_with_stdio(false) 37 38 typedef vector<int> VI; 39 typedef vector<string> VS; 40 typedef vector<double> VD; 41 typedef long long int64; 42 43 const double eps = 1e-12; 44 const double PI = atan(1.0)*4; 45 const int maxint = 2139062143; 46 47 inline int MyMod( int a , int b ) {a = a % b;if (a < 0) a += b; return a;} 48 49 class GForce 50 { 51 public: 52 double avgAccel(int p, vector <int> a, vector <int> t){ 53 int n = SZ (a); 54 VD spd; 55 spd.clear(); 56 for (int i = 0; i < n-1; ++ i){ 57 spd.PB(a[i] + 0.0); 58 double now = a[i] + 0.0; 59 double cnt = (a[i+1] - a[i] + eps) / (t[i+1] - t[i] + eps); 60 for (int j = t[i] + 1; j < t[i+1]; ++ j) 61 now += cnt, spd.PB (now); 62 } 63 spd.PB (a[n-1]); 64 65 double tmp = 0.0; 66 for (int i = 0; i < p; ++ i) 67 tmp += (spd[i] + spd[i+1]) / 2.0; 68 double ans = tmp; 69 for (int i = 1; i < spd.size() - p; ++ i){ 70 tmp -= (spd[i] + spd[i-1]) / 2.0; 71 tmp += (spd[i+p] + spd[i+p-1]) / 2.0; 72 ans = max (ans, tmp); 73 74 double del = (spd[i] + spd[i-1]) / 2.0; 75 double add = (spd[i+p] + spd[i+p-1]) / 2.0; 76 double haf = tmp + (3*spd[i]/8.0 + spd[i-1] / 8.0) - (3*spd[i+p]/8.0 + spd[i+p-1]/8.0); 77 ans = max (ans, haf); 78 if(fabs(del - add) > eps){ 79 del = (spd[i] - spd[i-1]) / 2.0; 80 add = (spd[i+p] - spd[i+p-1]) / 2.0; 81 double x = ((spd[i] - spd[i+p]) / 2.0) / (del - add); 82 if (x > eps && x < 1- eps){ 83 double y = ((spd[i]*spd[i+p-1] - spd[i+p]*spd[i-1]) / 2.0) / (del - add); 84 double xy = tmp + (spd[i] + y) * x / 2.0 - (y + spd[i+p]) * x / 2.0; 85 ans = max (ans, xy); 86 } 87 } 88 } 89 90 double ret = ans / (p + 0.0); 91 return (ret); 92 } 93 };
SRM 150
DIV2 1100pt
题意：有一个有限大小的平面和一个球，平面上有两种障碍物，第一种障碍物被球撞后会消失，第二种不会。球撞两种障碍物均会反弹，速度不变。给出平面大小和障碍物分布，问球从左上角一某一初速度开始，需要多少时间能让平面上的第一类障碍物全部消失。
解法：模拟。
tag: simulate
1 #line 10 "BrickByBrick.cpp" 2 #include <cstdlib> 3 #include <cctype> 4 #include <cstring> 5 #include <cstdio> 6 #include <cmath> 7 #include <algorithm> 8 #include <vector> 9 #include <iostream> 10 #include <sstream> 11 #include <set> 12 #include <queue> 13 #include <fstream> 14 #include <numeric> 15 #include <iomanip> 16 #include <bitset> 17 #include <list> 18 #include <stdexcept> 19 #include <functional> 20 #include <string> 21 #include <utility> 22 #include <map> 23 #include <ctime> 24 #include <stack> 25 26 using namespace std; 27 28 #define CLR(x) memset(x, 0, sizeof(x)) 29 #define PB push_back 30 #define SZ(v) ((int)(v).size()) 31 #define out(x) cout<<#x<<":"<<(x)<<endl 32 #define tst(a) cout<<#a<<endl 33 #define CINBEQUICKER std::ios::sync_with_stdio(false) 34 35 typedef vector<int> VI; 36 typedef vector<string> VS; 37 typedef vector<double> VD; 38 typedef long long int64; 39 40 const double eps = 1e-8; 41 const double PI = atan(1.0)*4; 42 const int maxint = 2139062143; 43 44 inline int MyMod( int a , int b ) {a = a % b;if (a < 0) a += b; return a;} 45 46 //drt 47 //1 0 2 48 //0 0 0 49 //3 0 4 50 struct STA{ 51 int x, y; 52 int drt; 53 int time; 54 void clr(){ 55 x = 1; y = 0; time = 0; 56 } 57 }; 58 59 int an[150][150]; 60 int xmax, ymax, tot; 61 int ans; 62 63 inline int change(char x) 64 { 65 if (x == '.') return 0; 66 if (x == 'B'){ 67 ++ tot; 68 return 1; 69 } 70 if (x == '#') return 2; 71 } 72 73 void init(VS m) 74 { 75 CLR (an); 76 tot = 0; 77 ymax = SZ (m); 78 for (int i = 0; i < ymax; ++ i){ 79 string s = m[i]; 80 xmax = SZ (s); 81 for (int j = 0; j < xmax; ++ j) 82 an[2*i+1][2*j+1] = change(s[j]); 83 } 84 } 85 86 void move(int& x, int& y, int d, int& t) 87 { 88 ++ t; 89 if (d == 1) -- x, -- y; 90 if (d == 2) ++ x, -- y; 91 if (d == 3) -- x, ++ y; 92 if (d == 4) ++ x, ++ y; 93 } 94 95 void D(int& x, int& y, int& d) 96 { 97 if (!(x & 1)){ 98 if (d & 1) 99 -- x, d = (d == 1 ? 2 : 4); 100 else 101 ++ x, d = (d == 2 ? 1 : 3); 102 return ; 103 } 104 if (d < 3) -- y, d = (d == 1 ? 3 : 4); 105 else ++ y, d = (d == 3 ? 1 : 2); 106 } 107 108 int solve() 109 { 110 STA sta; sta.clr(); 111 sta.drt = 4; 112 int num = 0; 113 while (sta.time < 1000000 && sta.time < ans){ 114 move(sta.x, sta.y, sta.drt, sta.time); 115 116 int x = sta.x, y = sta.y; 117 int dtmp = sta.drt; 118 D(x, y, sta.drt); 119 if (x >= 0 && x <= 2*xmax && y >= 0 && y <= 2*ymax){ 120 if (an[y][x] == 0) sta.drt = dtmp; 121 if (an[y][x] == 1) an[y][x] = 0, ++ num; 122 } 123 if (num == tot){ 124 return sta.time; 125 } 126 } 127 return -1; 128 } 129 130 class BrickByBrick 131 { 132 public: 133 int timeToClear(vector <string> m){ 134 init (m); 135 136 ans = maxint; 137 138 return (solve()); 139 } 140 };
DIV1 500pt
题意：给一个用‘A’-‘Z’表示颜色的字符串s，要求将一空白字符串变成给定字符串，（每次变化能将任意个连续位置的字符均变为某个任意的字符）问最少变化多少次。比如给定ABGBA，第一次将空白字符串变为AAAAA，第二次变为ABBBA，第三次变为ABGBA，答案为3。(s.size() <= 50)
解法：设置数组，m[][][]。left位置到left+size-1位置此时颜色均为c，将他们变化为给定字符串对应部分所需要最少的变化次数为m[left][size][c]。
状态转移方程：m[][0][] = 0,
if (color == s[l]) m[left][size][c] = m[left+1][size-1][color];
else m[left][size][color] = min {1 + m[left][i][color] + m[left+i][size-i][k] | 0 < i <= size, 'A' <= k <= 'Z'}
由于要用到记忆化搜索的思想，我又想写成DP，所以感觉写成了四不像- -
tag:dp, memoization, good
1 #line 10 "StripePainter.cpp" 2 #include <cstdlib> 3 #include <cctype> 4 #include <cstring> 5 #include <cstdio> 6 #include <cmath> 7 #include <algorithm> 8 #include <vector> 9 #include <iostream> 10 #include <sstream> 11 #include <set> 12 #include <queue> 13 #include <fstream> 14 #include <numeric> 15 #include <iomanip> 16 #include <bitset> 17 #include <list> 18 #include <stdexcept> 19 #include <functional> 20 #include <string> 21 #include <utility> 22 #include <map> 23 #include <ctime> 24 #include <stack> 25 26 using namespace std; 27 28 #define CLR(x) memset(x, 0, sizeof(x)) 29 #define PB push_back 30 #define SZ(v) ((int)(v).size()) 31 #define out(x) cout<<#x<<":"<<(x)<<endl 32 #define tst(a) cout<<#a<<endl 33 #define CINBEQUICKER std::ios::sync_with_stdio(false) 34 35 typedef vector<int> VI; 36 typedef vector<string> VS; 37 typedef vector<double> VD; 38 typedef long long int64; 39 40 const double eps = 1e-8; 41 const double PI = atan(1.0)*4; 42 const int maxint = 2139062143; 43 44 inline int MyMod( int a , int b ) {a = a % b;if (a < 0) a += b; return a;} 45 46 string s; 47 int m[55][55][100]; 48 49 int ask(int l, int n, char c) 50 { 51 int& a = m[l][n][(int)c]; 52 53 if (!n) return 0; 54 55 if (a != -1) return a; 56 a = maxint; 57 58 if (c == s[l]){ 59 a = ask (l+1, n-1, c); 60 return a; 61 } 62 63 for (int i = 1; i < n; ++ i) 64 for (int j = 'A'; j <= 'Z'; ++ j) 65 a = min (a, 1 + ask(l, i, s[l]) + ask(l+i, n-i, c)); 66 a = min (a, 1 + ask(l, n, s[l])); 67 return a; 68 } 69 70 class StripePainter 71 { 72 public: 73 int minStrokes(string stripes){ 74 s.clear(); 75 s = stripes; 76 memset (m, -1, sizeof (m)); 77 78 int size = SZ (s); 79 for (int i = 0; i < size; ++ i) 80 for (int j = 0; j + i <= size; ++ j) 81 for (int k = 'A'-1; k <= 'Z'; ++ k) 82 m[i][j][k] = 1 + ask(i, j, (char)k); 83 84 return (m[0][size][(int)s[0]]); 85 } 86 };