UVA 10645Menu （dp）

题意:

师傅会做n<=50种菜,供应d<=21天,每个菜都有花费和价值,师傅只有w<=100块钱
如果连续两天一样的菜,第二个价值减半,连续三天或以上,三天以后的都不算价值
求最大价值,打印一个可行方案

分析:

dp[i][w][today][yesterday]:=i天w花费,今天啥菜,昨天啥菜
暴力转移即可,然后从终止状态暴力搜出一组解

代码:

////  Created by TaoSama on 2015-11-07//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int d, n, m, w[55], v[55];double dp[22][105][55][55]; //k m today yesterday#define rep(i, a, b) for(int i = a; i <= b; ++i)void getMax(double& x, double y) {    x = max(x, y);}const double EPS = 1e-8;int sgn(double x) {    return x < -EPS ? -1 : x < EPS ? 0 : 1;}void print(int day, int cost, int today, int yesterday) {    if(day == 2) {        printf("%d %d ", yesterday, today);        return;    }    double cur = dp[day][cost][today][yesterday];    for(int i = 1; i <= n; ++i) {        double nxt = dp[day - 1][cost - w[today]][yesterday][i];        if(sgn(nxt) == 0) continue;        if(today == yesterday) {            if(yesterday != i) {                if(sgn(cur - (nxt + v[today] * 0.5)) == 0) {                    print(day - 1, cost - w[today], yesterday, i);                    printf("%d ", today);                    return;                }            } else if(sgn(cur - nxt) == 0) {                print(day - 1, cost - w[today], yesterday, i);                printf("%d ", today);                return;            }        } else if(sgn(cur - (nxt + v[today])) == 0) {            print(day - 1, cost - w[today], yesterday, i);            printf("%d ", today);            return;        }    }}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int kase = 0;    while(scanf("%d%d%d", &d, &n, &m) == 3 && (d || n || m)) {        w[0] = INF;        for(int i = 1; i <= n; ++i) scanf("%d%d", w + i, v + i);        if(kase++) puts("");        if(d == 1) {            int maxv = 0, ans = 0;            for(int i = 1; i <= n; ++i) if(w[i] <= m) maxv = max(maxv, v[i]);            for(int i = 1; i <= n; ++i)                if(v[i] == maxv && w[i] <= m && w[i] < w[ans]) ans = i;            if(maxv == 0) puts("0.0");            else printf("%.1f\n%d\n", 1.0 * maxv, ans);            continue;        }        memset(dp, 0, sizeof dp);        for(int i = 1; i <= n; ++i) {            for(int j = 1; j <= n; ++j) {                int weight = w[i] + w[j];                dp[2][weight][i][j] = i == j ? 1.5 * v[i] : v[i] + v[j];            }        }        rep(i, 3, d) rep(j, 1, n) rep(k, w[j], m) {            rep(x, 1, n) {                rep(y, 1, n) {                    double cur = dp[i - 1][k - w[j]][x][y];                    if(sgn(cur) == 0) continue;                    if(j == x) {                        if(y != x) cur += v[j] * 0.5;                    } else cur += v[j];                    getMax(dp[i][k][j][x], cur);                }            }        }        double maxv = 0;        rep(i, 0, m) rep(x, 1, n) rep(y, 1, n) getMax(maxv, dp[d][i][x][y]);        int cost = INF, today, yesterday;        rep(i, 0, m) {            rep(x, 1, n) rep(y, 1, n) {                if(sgn(dp[d][i][x][y] - maxv) == 0) {                    cost = i;                    today = x;                    yesterday = y;                }            }            if(cost != INF) break;        }        if(cost == 0) {puts("0.0"); continue;}        printf("%.1f\n", maxv);        print(d, cost, today, yesterday);        puts("");    }    return 0;}