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这道题目的意思就是排两排书，下面这排只能竖着放，上面这排可以平着放，使得宽度最小
根据题意可以得出一个结论，放上这排书的Width 肯定会遵照从小到大的顺序放上去的
Because the total thickness of vertical books is fixed it‘s good to calculate the minimum possible total width of horizontal books.
那么只需要模拟一遍放书的过程即可，不会TLE
不过正统解法是Dp
Dp Source Code:
//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler #include <stdio.h> #include <iostream> #include <fstream> #include <cstring> #include <cmath> #include <stack> #include <string> #include <map> #include <set> #include <list> #include <queue> #include <vector> #include <algorithm> #define Max(a,b) (((a) > (b)) ? (a) : (b)) #define Min(a,b) (((a) < (b)) ? (a) : (b)) #define Abs(x) (((x) > 0) ? (x) : (-(x))) #define MOD 1000000007 #define pi acos(-1.0) using namespace std; typedef long long ll ; typedef unsigned long long ull ; typedef unsigned int uint ; typedef unsigned char uchar ; template<class T> inline void checkmin(T &a,T b){if(a>b) a=b;} template<class T> inline void checkmax(T &a,T b){if(a<b) a=b;} const double eps = 1e-7 ; const int N = 210 ; const int M = 1100011*2 ; const ll P = 10000000097ll ; const int MAXN = 10900000 ; const int INF = 0x3f3f3f3f ; int dp[101][210][301]; int main(){ std::ios::sync_with_stdio(false); int i, j, t, k, u, v, x, y, numCase = 0; int n, a, b; dp[0][0][0] = 1; cin >> n; int cur = 0; for(i = 0; i < n; ++i){ cin >> a >> b; for(j = 0; j < cur + 1; ++j){ for(k = 0; k < 201; ++k){ dp[i + 1][j + a][k] |= dp[i][j][k]; dp[i + 1][j][k + b] |= dp[i][j][k]; } } cur += a; } for(i = 0; i < 201; ++i){ for(j = 0; j < i + 1; ++j){ if(dp[n][i][j]){ cout << i << endl; return 0; } } } return 0; }
My Source Code:
//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler #include <stdio.h> #include <iostream> #include <fstream> #include <cstring> #include <cmath> #include <stack> #include <string> #include <map> #include <set> #include <list> #include <queue> #include <vector> #include <algorithm> #define Max(a,b) (((a) > (b)) ? (a) : (b)) #define Min(a,b) (((a) < (b)) ? (a) : (b)) #define Abs(x) (((x) > 0) ? (x) : (-(x))) #define MOD 1000000007 #define pi acos(-1.0) using namespace std; typedef long long ll ; typedef unsigned long long ull ; typedef unsigned int uint ; typedef unsigned char uchar ; template<class T> inline void checkmin(T &a,T b){if(a>b) a=b;} template<class T> inline void checkmax(T &a,T b){if(a<b) a=b;} const double eps = 1e-7 ; const int N = 210 ; const int M = 1100011*2 ; const ll P = 10000000097ll ; const int MAXN = 10900000 ; const int INF = 0x3f3f3f3f ; int a[10][122]; int one, two, n; int main(){ std::ios::sync_with_stdio(false); int i, j, t, k, u, v, x, y, numCase = 0; cin >> n; for(u = 0; u < n; ++u){ cin >> x >> y; if(x == 1){ a[1][one++] = y; } else{ a[2][two++] = y; } } sort(a[1], a[1] + one); sort(a[2], a[2] + two); int ans = INF; for(i = 0; i <= one; ++i){ for(j = 0; j <= two; ++j){ int cur = 0; for(k = 0; k < one - i; ++k){ cur += a[1][k]; } for(k = 0; k < two - j; ++k){ cur += a[2][k]; } if((i + 2 * j) < ans && (i + 2 * j) >= cur){ ans = i + 2 * j; } } } cout << ans << endl; return 0; }
CodeForces 294B Shaass and Bookshelf 【规律 & 模拟】或【Dp】
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原文地址：http://www.cnblogs.com/wushuaiyi/p/4314285.html
