大意: n个红黑卡, 每天可以选择领取一块红币一块黑币, 或者买一张卡, 第$i$张卡的花费红币数$max(r_i-A,0)$, 花费黑币数$max(b_i-B,0)$, A为当前红卡数, B为当前黑卡数, 求买完所有卡最少天数.
这题挺巧妙的, 刚开始看花费的范围太大一直在想怎么贪心...
实际上注意到减费最多只有120, 可以按照减费进行dp即可
这题CF大神的最优解写了个模拟退火ORZ
#include <iostream> #include <algorithm> #include <cstdio> #include <math.h> #include <set> #include <map> #include <queue> #include <string> #include <string.h> #include <bitset> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc (o<<1) #define rc (lc|1) #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl ‘\n‘ using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} //head const int N = 20; int n; char c[N]; int x[N], y[N]; int dp[1<<16][150]; void chkmax(int &a, int b) {a=max(a,b);} int main() { scanf("%d", &n); REP(i,0,n-1) scanf(" %c%d%d", c+i, x+i, y+i); memset(dp, 0xbc, sizeof dp); dp[0][0] = 0; int mx = (1<<n)-1; REP(i,0,mx-1) { int A = 0, B = 0; REP(k,0,n-1) if (i>>k&1) { if (c[k]==‘R‘) ++A; else ++B; } REP(k,0,n-1) if (!(i>>k&1)) { REP(j,0,120) if (dp[i][j]!=0xbcbcbcbc) { chkmax(dp[i^1<<k][j+min(x[k], A)],dp[i][j]+min(y[k], B)); } } } int ans = INF, X = 0, Y = 0; REP(i,0,n-1) X+=x[i], Y+=y[i]; REP(i,0,120) ans = min(ans, max(X-i, Y-dp[mx][i])); printf("%d\n", ans+n); }
原文地址:https://www.cnblogs.com/uid001/p/10623739.html
时间: 2024-11-07 23:41:15