SRM 719 Div 1 250 500

250：

题目大意：

在一个N行无限大的网格图里，每经过一个格子都要付出一定的代价。同一行的每个格子代价相同。 给出起点和终点，求从起点到终点的付出的最少代价。

思路：

最优方案肯定是从起点沿竖直方向走到某一行，然后沿水平方向走到终点那一列，然后再沿竖直方向走到终点那一行。

枚举是通过哪一行的格子从起点那列走到终点那列的，求个最小值就好了。

代码：

 // BEGIN CUT HERE  

 // END CUT HERE
 #line 5 "LongMansionDiv1.cpp"
 #include <vector>
 #include <list>
 #include <map>
 #include <set>
 #include <deque>
 #include <stack>
 #include <bitset>
 #include <algorithm>
 #include <functional>
 #include <numeric>
 #include <utility>
 #include <sstream>
 #include <iostream>
 #include <iomanip>
 #include <cstdio>
 #include <cmath>
 #include <cstdlib>
 #include <ctime>
 #include <cstring>
 using namespace std;  

 typedef long long ll;

 class LongMansionDiv1
 {
 public:
 ll F(int i, int j, vector <int> &t)
 {
     ll res = ;
     if (i > j) swap(i, j);
     for (int k = i; k <= j; ++k)
         res += t[k];
     return res;
 }
 long long minimalTime(vector <int> t, int sX, int sY, int eX, int eY)
 {
 //$CARETPOSITION$
     int n = t.size();
     if (sY > eY) swap(sX, eX), swap(sY, eY);
     ll ans = 1e18;
     for (int j = ; j < n; ++j)
     {
         ans = min(ans, F(sX, j, t) + 1ll * t[j] * (eY - sY + ) + F(j, eX, t) - t[j] - t[j]);
     }
     return ans;
 }  

 // BEGIN CUT HERE
     public:
     void run_test(int Case) { if ((Case == -) || (Case == )) test_case_0(); if ((Case == -) || (Case == )) test_case_1(); if ((Case == -) || (Case == )) test_case_2(); if ((Case == -) || (Case == )) test_case_3(); if ((Case == -) || (Case == )) test_case_4(); }
     private:
     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(); }
     void verify_case(int Case, const long long &Expected, const long long &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
     void test_case_0() { int Arr0[] = {, , }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; int Arg2 = ; int Arg3 = ; int Arg4 = ; long long Arg5 = 29LL; verify_case(, Arg5, minimalTime(Arg0, Arg1, Arg2, Arg3, Arg4)); }
     void test_case_1() { int Arr0[] = {, , }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; int Arg2 = ; int Arg3 = ; int Arg4 = ; long long Arg5 = 15LL; verify_case(, Arg5, minimalTime(Arg0, Arg1, Arg2, Arg3, Arg4)); }
     void test_case_2() { int Arr0[] = {, , , , , , }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; int Arg2 = ; int Arg3 = ; int Arg4 = ; long long Arg5 = 1016LL; verify_case(, Arg5, minimalTime(Arg0, Arg1, Arg2, Arg3, Arg4)); }
     void test_case_3() { int Arr0[] = {, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; int Arg2 = ; int Arg3 = ; int Arg4 = ; long long Arg5 = 293443080673LL; verify_case(, Arg5, minimalTime(Arg0, Arg1, Arg2, Arg3, Arg4)); }
     void test_case_4() { int Arr0[] = {}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; int Arg2 = ; int Arg3 = ; int Arg4 = ; long long Arg5 = 1LL; verify_case(, Arg5, minimalTime(Arg0, Arg1, Arg2, Arg3, Arg4)); }

 // END CUT HERE

 };  

 // BEGIN CUT HERE
 int main()
 {
 LongMansionDiv1 ___test;
 ___test.run_test(-);
 system("pause");
 }
 // END CUT HERE  

---

500：

题目大意：

给出一棵以0为根的树，从根出发，走过一些节点。 每个节点有一个得分，可能正可能负，可以重复经过节点，但是只有第一次经过会改变当前得分。 如果当前得分为负，会马上变成0.

求最大得分。

比赛的时候没有想出来，好菜菜QAQ。

思路：

这道题主要是 “如果当前得分为负，会马上变成0” 这个地方不好处理。

考虑最后一次得分由负变成0的时刻。这个时候访问过的节点也组成以0为根的树，可以认为这棵树上的节点的得分都是0.

因此可以把问题转化一下：

可以把包含节点0的一个联通块内的节点得分都变成0， 然后再求一个包含节点0的联通块得分和最大。

用A，B两个数组来做DP。

A[x]表示考虑以x为根的子树，且x必须取的最大值。

B[x]表示考虑以x为根的子树，且允许把x的权值变成0（相当于允许不取x）的最大值。

A[x] =  max(val[x] + sum(A[sons of x]), 0)

B[x] =  max(sum(B[sons of x], A[x]))

代码：

 // BEGIN CUT HERE  

 // END CUT HERE
 #line 5 "OwaskiAndTree.cpp"
 #include <vector>
 #include <list>
 #include <map>
 #include <set>
 #include <deque>
 #include <stack>
 #include <bitset>
 #include <algorithm>
 #include <functional>
 #include <numeric>
 #include <utility>
 #include <sstream>
 #include <iostream>
 #include <iomanip>
 #include <cstdio>
 #include <cmath>
 #include <cstdlib>
 #include <ctime>
 #include <cstring>
 using namespace std;
 class OwaskiAndTree
 {
 public:
 int maximalScore(vector <int> parent, vector <int> pleasure)
 {
 //$CARETPOSITION$
     int n = pleasure.size();
     long long a[] = {}, b[] = {};
     for (int i = n - ; i >= ; --i)
     {
         a[i] += pleasure[i];
         a[i] = max(a[i], 0ll);
         b[i] = max(b[i], a[i]);

         if (i == ) continue;
         int u = parent[i - ];
         a[u] += max(a[i], 0ll);
         b[u] += b[i];
     }
     return b[];
 }  

 // BEGIN CUT HERE
     public:
     void run_test(int Case) { if ((Case == -) || (Case == )) test_case_0(); if ((Case == -) || (Case == )) test_case_1(); if ((Case == -) || (Case == )) test_case_2(); if ((Case == -) || (Case == )) test_case_3(); if ((Case == -) || (Case == )) test_case_4(); }
     private:
     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(); }
     void verify_case(int Case, const int &Expected, const int &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
     void test_case_0() { int Arr0[] = {, , , , , , , , }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arr1[] = {, , -, -, -, -, , , , }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); int Arg2 = ; verify_case(, Arg2, maximalScore(Arg0, Arg1)); }
     void test_case_1() { int Arr0[] = {, , , }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arr1[] = {, , , -, -}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); int Arg2 = ; verify_case(, Arg2, maximalScore(Arg0, Arg1)); }
     void test_case_2() { int Arr0[] = {, , , , , , , }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arr1[] = {, , -, -, , , , -, }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); int Arg2 = ; verify_case(, Arg2, maximalScore(Arg0, Arg1)); }
     void test_case_3() { int Arr0[] = {, , , , , , , , , , , , , , , , , , }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arr1[] = {-, , , , -, , -, -, , -, , -, , , -, , , -, , -}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); int Arg2 = ; verify_case(, Arg2, maximalScore(Arg0, Arg1)); }
     void test_case_4() { int Arr0[] = {}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arr1[] = {-}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); int Arg2 = ; verify_case(, Arg2, maximalScore(Arg0, Arg1)); }

 // END CUT HERE

 };  

 // BEGIN CUT HERE
 int main()
 {
 OwaskiAndTree ___test;
 ___test.run_test(-);
 system("pause");
 }
 // END CUT HERE