HDU 2159 FATE (二维背包)

Posted 2020-08-13 dwtfukgv

题意：中文题。

析：dp[i][j] 已经杀了 i 个怪兽，已经用了 j 体积，所能获得的最大经验值，这个和一维的差不多，只是加一维而已。

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>
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

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 = 1e2 + 5;
const LL mod = 10000000000007;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
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 dp[maxn][maxn];
int a[maxn], c[maxn];

int main(){
    int q, s;
    while(scanf("%d %d %d %d", &n, &m, &q, &s) == 4){
        for(int i = 0; i < q; ++i)  scanf("%d %d", a+i, c+i);

        memset(dp, 0, sizeof dp);
        int ans = -1;
        for(int i = 1; i <= s; ++i){
            for(int k = 0; k < q; ++k){
                for(int j = c[k]; j <= m; ++j){
                    dp[i][j] = Max(dp[i][j], dp[i-1][j-c[k]] + a[k]);
                    if(dp[i][j] >= n){  ans = Max(ans, m-j);  break; }
                }
            }
        }

        printf("%d\n", ans);
    }
    return 0;
}