2、0-1背包问题，边界问题稍微麻烦一些。
1 #include <iostream> 2 #include <string> 3 #include <vector> 4 #include <cstdlib> 5 #include <map> 6 #include <algorithm> 7 #include <stack> 8 #include <queue> 9 #include <cmath> 10 using namespace std; 11 int dp[105][105];//dp[i][j],从第i个card开始，后边剩余j张牌的情况下最大的damage 12 class SpellCards { 13 public: 14 int maxDamage(vector<int> level, vector<int> damage) { 15 int dsz = damage.size(); 16 int lsz = level.size(); 17 int res = 0; 18 19 for (int id = 0; id < dsz; id++) { 20 rotate(level.begin(), level.begin() + 1, level.end()); 21 rotate(damage.begin(), damage.begin() + 1, damage.end()); 22 memset(dp, -1, sizeof(dp)); 23 24 if (level[lsz - 1] == 1) {//初始化倒数第一个 25 dp[lsz - 1][0] = damage[lsz - 1]; 26 } 27 dp[lsz - 1][1] = 0; 28 29 for (int i = lsz - 2; i > -1; i--) { //倒数第二个个位置开始 30 for (int j = 0; j < lsz - i + 1; j++) {//后边还剩下j个可以用 31 if (0 == j) { //初始化dp的值，表示不使用当前的card 32 dp[i][j] = -1; 33 } else { 34 dp[i][j] = dp[i + 1][j - 1]; 35 } 36 37 int use_need = level[i] - 1; //使用之后需要的个数 38 if (dp[i + 1][j + use_need] != -1) {//如果可以使用 39 dp[i][j] = max(dp[i][j], 40 dp[i + 1][j + use_need] + damage[i]); 41 res = max(res, dp[i][j]); 42 } 43 } 44 } 45 } 46 return res; 47 } 48 };
递归版本
1 #include <iostream> 2 #include <string> 3 #include <vector> 4 #include <cstdlib> 5 #include <map> 6 #include <algorithm> 7 #include <stack> 8 #include <queue> 9 #include <cmath> 10 using namespace std; 11 int dp[105][105];//dp[i][j],从第i个card开始，后边剩余j张牌的情况下最大的damage 12 class SpellCards { 13 public: 14 int rec(vector<int>& level, vector<int>& damage,int cur,int need,int total){ 15 int &res=dp[cur][need]; 16 if(-1==res){ 17 if((cur+need)==total){ 18 res=0; 19 }else{ 20 res=rec(level,damage,cur+1,max(need-1,0),total);//不使用 21 if((cur+need+level[cur])<=total){//使用 22 res=max(res,damage[cur]+rec(level,damage,cur+1,need+level[cur]-1,total)); 23 } 24 } 25 } 26 return res; 27 } 28 int maxDamage(vector<int> level, vector<int> damage) { 29 int dsz = damage.size(); 30 int lsz = level.size(); 31 int res = 0; 32 33 for (int id = 0; id < dsz; id++) { 34 rotate(level.begin(), level.begin() + 1, level.end()); 35 rotate(damage.begin(), damage.begin() + 1, damage.end()); 36 memset(dp, -1, sizeof(dp)); 37 res = max(res, rec(level,damage,0,0,lsz)); 38 } 39 return res; 40 } 41 };
最高效方法：http://www.cppblog.com/hanfei19910905/archive/2012/12/09/196119.aspx
1 #include <iostream> 2 #include <string> 3 #include <vector> 4 #include <cstdlib> 5 #include <map> 6 #include <algorithm> 7 #include <stack> 8 #include <queue> 9 #include <cmath> 10 using namespace std; 11 int dp[55]; 12 //dp[i]当前有i张牌的情况下的最大的damage 13 int work(vector<int> a, vector<int> d) { 14 int n = a.size(), ans = 0; 15 memset(dp, -1, sizeof(dp)); 16 dp[0] = 0; 17 for (int j = n - 1; j > -1; j--) { 18 for (int i = n; i; i--) 19 if (i - a[j] >= 0) { 20 int v = i - a[j]; 21 if (dp[v] == -1) 22 continue; 23 dp[i] = max(dp[i], dp[v] + d[j]); 24 ans = max(ans, dp[i]); 25 } 26 } 27 return ans; 28 } 29 class SpellCards { 30 public: 31 int maxDamage(vector<int> level, vector<int> damage) { 32 return work(level, damage); 33 } 34 };