lightoj - 1169 - Monkeys on Twin Tower - dp

来源：互联网发布：软件测评师好考吗编辑：程序博客网时间：2024/05/08 20:16

题意：有两座塔，有n层，每层有一个水果，吃掉水果的时间分别时a[i], b[i]，从i到i + 1层有两种方式，一种是直接上去，不消耗时间，另一种是跳到另一座塔，消耗u[i]或v[i]的时间。要跳到最高层，求最小时间。

题解：每层有两种决策。

#include <bits/stdc++.h>//#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define ll long long#define SZ(x) ((int)(x).size()) #define ALL(v) (v).begin(), (v).end()#define foreach(i, v) for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++ i)#define reveach(i, v) for (__typeof((v).rbegin()) i = (v).rbegin(); i != (v).rend(); ++ i) #define REP(i,a,n) for ( int i=a; i<int(n); i++ )#define FOR(i,a,n) for ( int i=n-1; i>= int(a);i-- )#define lson rt<<1, L, m#define rson rt<<1|1, m, Rtypedef pair<int, int> pii;typedef pair<ll, ll> pll;#define mp(x, y) make_pair(x, y)#define pb(x) push_back(x)#define fi first#define se second#define CLR(a, b) memset(a, b, sizeof(a))#define Min(a, b) a = min(a, b)#define Max(a, b) a = max(a, b)const int maxn = 1e3 + 7;const int maxm = 5e3 + 7;int T;int kase;int n;int a[maxn];int b[maxn];int u[maxn];int v[maxn];int dp[maxn][2];int solve(){    dp[0][0] = a[0], dp[0][1] = b[0];    REP(i, 1, n){        dp[i][0] = min(dp[i-1][0], dp[i-1][1] + v[i-1]) + a[i];        dp[i][1] = min(dp[i-1][1], dp[i-1][0] + u[i-1]) + b[i];    }    return min(dp[n-1][0], dp[n-1][1]);}int main(){#ifdef ac    freopen("in.txt","r",stdin);#endif    //freopen("out.txt","w",stdout);    scanf("%d", &T);    while(T--){        scanf("%d", &n);        REP(i, 0, n) scanf("%d", &a[i]);        REP(i, 0, n) scanf("%d", &b[i]);        REP(i, 0, n-1) scanf("%d", &u[i]);        REP(i, 0, n-1) scanf("%d", &v[i]);        printf("Case %d: %d\n", ++ kase, solve());    }    return 0;}

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