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UVaLive 6853 Concert Tour (DP)

DP UVALive
2023-09-11 14:17:18 时间

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：

#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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