ZOJ 2972 Hurdles of 110m 【DP 背包】

一共有N段过程，每段过程里可以选择 快速跑、 匀速跑 和 慢速跑

对于快速跑会消耗F1 的能量， 慢速跑会集聚F2的能量

选手一开始有M的能量，即能量上限

求通过全程的最短时间

定义DP[i][j] 为跨越第 i 个栏，剩余 j 点能量

动态转移方程 

dp[i][j] = min(dp[i][j], dp[i-1][j-F1]+T1)　　  (Fast Mode)

dp[i][j] = min(dp[i][j], dp[i-1][j]+T2) 　　　　(Normal Mode)

dp[i][j] = min(dp[i][j], dp[i-1][j+F2]+T3) 　　(Slow Mode)

Source Code:

//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler
#include <stdio.h>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cmath>
#include <stack>
#include <string>
#include <map>
#include <set>
#include <list>
#include <queue>
#include <vector>
#include <algorithm>
#define Max(a,b) (((a) > (b)) ? (a) : (b))
#define Min(a,b) (((a) < (b)) ? (a) : (b))
#define Abs(x) (((x) > 0) ? (x) : (-(x)))
#define MOD 1000000007
#define pi acos(-1.0)

using namespace std;

typedef long long           ll      ;
typedef unsigned long long  ull     ;
typedef unsigned int        uint    ;
typedef unsigned char       uchar   ;

template<class T> inline void checkmin(T &a,T b){if(a>b) a=b;}
template<class T> inline void checkmax(T &a,T b){if(a<b) a=b;}

const double eps = 1e-7      ;
const int N = 22             ;
const int M = 1100011*2      ;
const ll P = 10000000097ll   ;
const int MAXN = 100000      ;
const int INF = 0x3f3f3f3f   ;
const int MAX = 10500        ;

struct sc{
    int t1, t2, t3, f1, f2;
}a[120];

int n, m;
int dp[120][120];

int main(){
    std::ios::sync_with_stdio(false);
    int i, j, t, k, l, u, v, x, y, numCase = 0;
    cin >> t;
    while(t--){
        cin >> n >> m;
        for(i = 0; i < n; ++i){
            cin >> a[i].t1 >> a[i].t2 >> a[i].t3 >> a[i].f1 >> a[i].f2;
        }
        memset(dp, 0x3f, sizeof(dp));
        dp[0][m] = 0;
        for(i = 0; i < n; ++i){
            for(j = 0; j <= m; ++j){
                if(j >= a[i].f1){
                    checkmin(dp[i + 1][j - a[i].f1], dp[i][j] + a[i].t1);
                }
                if(j + a[i].f2 > m){
                    checkmin(dp[i + 1][m], dp[i][j] + a[i].t3);
                } else{
                    checkmin(dp[i + 1][j + a[i].f2], dp[i][j] + a[i].t3);
                }
                checkmin(dp[i + 1][j], dp[i][j] + a[i].t2);
            }
        }
        int ans = INF;
        for(i = 0; i <= m; ++i){
            checkmin(ans, dp[n][i]);
        }
        cout << ans << endl;
    }

    return 0;
}