UVaLive 6853 Concert Tour (DP)
2016-09-14 17:28
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题意:给定 n 个城市,m 个月,表示要在这 n 个城市连续 m 个月开演唱会,然后给定每个月在每个城市开演唱会能获得的利润,然后就是演唱会在不同城市之间调动所要的费用,
问你,怎么安排这 n 个演唱会是最优的。
析:很明显的一个DP题,并且也不难,用dp[i][j] 表示在第 i 个月,在第 j 个城市开演唱会,是最优的。那么状态转移方程也就出来了
dp[i][j] = Max(dp[i][j], dp[i-1][k]-f[k][j]+p[j][i]);
代码如下:
问你,怎么安排这 n 个演唱会是最优的。
析:很明显的一个DP题,并且也不难,用dp[i][j] 表示在第 i 个月,在第 j 个城市开演唱会,是最优的。那么状态转移方程也就出来了
dp[i][j] = Max(dp[i][j], dp[i-1][k]-f[k][j]+p[j][i]);
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const LL LNF = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e3 + 100; const int mod = 1e9 + 7; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline int Min(int a, int b){ return a < b ? a : b; } inline int Max(int a, int b){ return a > b ? a : b; } inline LL Min(LL a, LL b){ return a < b ? a : b; } inline LL Max(LL a, LL b){ return a > b ? a : b; } inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int p[105][55]; int f[105][105]; int dp[55][105]; int main(){ int T; cin >> T; while(T--){ scanf("%d %d", &n, &m); for(int i = 1; i <= n; ++i) for(int j = 1; j <= m; ++j) scanf("%d", p[i]+j); for(int i = 1; i <= n; ++i) for(int j = 1; j <= n; ++j) scanf("%d", f[i]+j); memset(dp, 0, sizeof dp); for(int i = 1; i <= n; ++i) dp[1][i] = p[i][1]; for(int i = 2; i <= m; ++i) for(int j = 1; j <= n; ++j) for(int k = 1; k <= n; ++k) dp[i][j] = Max(dp[i][j], dp[i-1][k]-f[k][j]+p[j][i]); int ans = 0; for(int i = 1; i <= n; ++i) ans = Max(ans, dp[m][i]); printf("%d\n", ans); } return 0; }
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