2010杭州賽區網絡預賽 1006 Fate Stay Night （DP）

The 35th ACM/ICPC Asia Regional Hangzhou Site —— Online Contest  1006  Fate Stay Night

題目鏈接 http://acm.hdu.edu.cn/showproblem.php?pid=3646  這裏還可以提交

這個題主要就是用DP，不過動態方程算是比較難想的

 // 1006 mnlm 1.0 #include<cstdio> #include<cstdlib> #include<cmath> #include<cstring> #include<vector> #include<algorithm> using namespace std; int k[100001]; int n; int m; int K; int b[10001]; struct DP { int ans; int blood; bool operator < (const DP &other) const { if (ans != other.ans) { return ans < other.ans; } else { return blood > other.blood; } } }dp[101][10001][2]; // dp[a][b][c]中的a表示用過了a次”Command Mantra“，b表示進行到第b個fire bird // c爲1 時表示進行到第b個fire bird的時候用了”Command Mantra“，0則沒有用 void Deal(int tt, DP & r) { bool flag = true; while (tt > 0 && r.ans < K) { if (tt > r.blood) { tt -= r.blood; r.ans++; r.blood = k[r.ans]; } else if (tt == r.blood) { tt = 0; r.ans++; r.blood = k[r.ans]; } else { if (flag) { r.blood -= tt; } tt = 0; } flag = false; } } int main() { while (1) { scanf("%d%d%d", &n, &m, &K); if (!n && !m && !K) { break; } int i; for (i = 0; i < n; i++) { scanf("%d", &b[i + 1]); } for (i = 0; i < K; i++) { scanf("%d", &k[i]); } int j; dp[0][0][0].ans = 0; dp[0][0][0].blood = k[0]; dp[0][0][1].ans = 0; dp[0][0][1].blood = k[0]; for (i = 1; i <= n; i++) { DP & r = dp[0][i][0]; r.ans = dp[0][i - 1][0].ans; r.blood = dp[0][i - 1][0].blood; Deal(b[i], r); dp[0][i][1].ans = 0; dp[0][i][1].blood = 0x7fffffff; } for (i = 1; i <= m; i++) { dp[i][0][0].ans = 0; dp[i][0][0].blood = k[0]; dp[i][0][1].ans = 0; dp[i][0][1].blood = k[0]; for (j = 1; j <= n; j++) { DP & p = dp[i][j - 1][0] < dp[i][j - 1][1] ? dp[i][j - 1][1] : dp[i][j - 1][0]; DP & q = dp[i][j][0]; q.ans = p.ans; q.blood = p.blood; Deal(b[j], q); DP & w = dp[i][j][1]; DP & e = dp[i - 1][j - 1][0] < dp[i - 1][j - 1][1] ? dp[i - 1][j - 1][1] : dp[i - 1][j - 1][0]; w.ans = e.ans; w.blood = e.blood; Deal(2 * b[j], w); } } if (dp[m][n][1] < dp[m][n][0]) { printf("%d/n", dp[m][n][0].ans); } else { printf("%d/n", dp[m][n][1].ans); } } }