HDU 5445 Food Problem、UVa 10163 Storage Keepers、POJ 3260 The Fewest Coins（两次dp）

Food Problem

题意: 给你n种食物，m种车，每种食物有三种属性能量值t，体积u，数量v。每种车有三个属性值容量x,价格y,数量z。

         问题是在能够达到至少p能量的要求下，最小花费为多少，若大于50000则输出TAT 

分析: 两次多重背包dp 先dp出至少p能量的最小体积 然后从50000花费再dp出体积 在满足之前的最小体积下找到答案

         为啥不体积dp出花费呢? - - 看看哪个数量级大就知道了 要选取合适的状态减少复杂度 233

         记得二进制优化233

代码:

////  Created by TaoSama on 2015-09-25//  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 n, m, p;int dp[60005];//1: dp[i][j]:= i desert j energy's minimum size//2: dp[i][j]:= i truck j cost's maximum sizeint main() {#ifdef LOCAL    freopen("in.txt", "r", stdin);//  freopen("out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    while(t--) {        scanf("%d%d%d", &n, &m, &p);        memset(dp, 0x3f, sizeof dp);        dp[0] = 0;        for(int i = 1; i <= n; ++i) {            int w, v, c; scanf("%d%d%d", &w, &v, &c);            for(int k = 1; c > 0; c -= k, k <<= 1) {                int mul = min(k, c);                for(int j = p + 100; j >= mul * w; --j)                    dp[j] = min(dp[j], dp[j - mul * w] + mul * v);            }        }        int V = INF;        for(int i = p; i <= p + 100; ++i) V = min(V, dp[i]);//      printf("V: %d\n", V);        memset(dp, 0, sizeof dp);        int ans = INF;        for(int i = 1; i <= m; ++i) {            int v, w, c; scanf("%d%d%d", &v, &w, &c);            for(int k = 1; c > 0; c -= k, k <<= 1) {                int mul = min(k, c);                for(int j = 50000; j >= mul * w; --j) {                    dp[j] = max(dp[j], dp[j - mul * w] + mul * v);                    if(dp[j] >= V) ans = min(ans, j);                }            }        }        if(ans == INF) puts("TAT");        else printf("%d\n", ans);    }    return 0;}

Storage Keepers

题意: 有n<100个仓库  m<=30个管理员  每个管理员有一个能力值P(接下来的一行有m个数，表示每个管理员的能力值)
          每个仓库只能由一个管理员看管，但是每个管理员可以看管k个仓库（但是这个仓库分配到的安全值只有p/k,k=0,1,...
          每个月公司都要给看管员工资，雇用的管理员的工资即为他们的能力值p和，求使每个仓库的安全值最高的前提下 使的工资总和最小。
          输出最大安全值，并且输出最少的花费。

分析: 两次类似于多重背包的dp 状态见代码 第一次dp出安全值 第二次再在这个状态下找到最少花费  注意初始化好好想想啊

代码:

////  Created by TaoSama on 2015-08-12//  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 n, m, p[35];int f[35][105], g[35][105];//dp[i][j]:= i people look after j storages max safe, min wageint main() {#ifdef LOCAL    freopen("in.txt", "r", stdin);//  freopen("out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    while(scanf("%d%d", &m, &n) == 2 && (n || m)) {        for(int i = 1; i <= n; ++i) scanf("%d", p + i);        memset(f, 0, sizeof f);        f[0][0] = INF; //initialization is so difficult!        for(int i = 1; i <= n; ++i) {            for(int j = 0; j <= m; ++j) {                f[i][j] = f[i - 1][j]; //zero exception                for(int k = 1; k <= p[i] && k <= j; ++k)                    f[i][j] = max(f[i][j], min(f[i - 1][j - k], p[i] / k));            }        }        int L = f[n][m];        if(!L) {printf("0 0\n"); continue;}        memset(g, 0x3f, sizeof g);        g[0][0] = 0;        for(int i = 1; i <= n; ++i) {            for(int j = 0; j <= m; ++j) {                g[i][j] = g[i - 1][j];                for(int k = 1; k <= p[i] && k <= j; ++k) {                    int s = p[i] / k;  //at least                    if(s >= f[n][m]) g[i][j] = min(g[i][j], g[i - 1][j - k] + p[i]);                }            }        }        int Y = g[n][m];        printf("%d %d\n", L, Y);    }    return 0;}

The Fewest Coins

题意: 给出钱币的种类数和总的购买值  然后给出每种钱币的价值与数量

         老板也是每种钱币都拥有  但是没有数量限制  购买东西的时候 价值超过给定价值的话 老板会找钱

         求最小的交流钱币的数量

分析: 两次dp啦, 第一次多重背包 求出用了多少硬币  然后第二次完全背包求出找了多少硬币 然后求个最小值

         最重要的是上界的处理。可以注意到，上界为maxw*maxw+m（maxw最大面额的纸币），也就是24400元。

         贴个证明:
         如果买家的付款数大于了maxw*maxw+m，即付硬币的数目大于了maxw，根据鸽笼原理，至少有两个的和对maxw取模的值相等

         也就是说，这部分硬币能够用更少的maxw来代替。证毕。
         真心不懂证明也没关系，记住结论就好了，真心不想记结论也没关系，数组开大一点就好了。

代码:

////////  Created by TaoSama on 2015-03-30//  Copyright (c) 2015 TaoSama. All rights reserved.//#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;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;const int N = 1e5 + 10;int n, t, dpp[25000], dpn[15000], c[105], v[105];int main() {#ifdef LOCAL    freopen("in.txt", "r", stdin);//freopen("out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    cin >> n >> t;    for(int i = 1; i <= n; ++i) cin >> v[i];    for(int i = 1; i <= n; ++i) cin >> c[i];    memset(dpp, 0x3f, sizeof dpp);    dpp[0] = 0;    for(int i = 1; i <= n; ++i) {        int k = 1;        while(c[i] > 0) {            int t = min(c[i], k);            for(int j = t + 120 * 120; j >= t * v[i]; --j)                dpp[j] = min(dpp[j], dpp[j - t * v[i]] + t);            c[i] -= k; k <<= 1;        }    }    /*for(int i = 0; i <= t + 120 * 120; ++i)        if(dpp[i] != INF) printf("dpp[%d]: %d\n", i, dpp[i]);*/    memset(dpn, 0x3f, sizeof dpn);    dpn[0] = 0;    for(int i = 1; i <= n; ++i)        for(int j = v[i]; j <= 120 * 120; ++j)            dpn[j] = min(dpn[j], dpn[j - v[i]] + 1);    /*for(int i = 0; i <= 120 * 120; ++i)        if(dpn[i] != INF) printf("dpn[%d]: %d\n", i, dpn[i]);*/    int ans = INF;    for(int i = 0; i <= 120 * 120; ++i)        ans = min(ans, dpp[t + i] + dpn[i]);    if(ans == INF) ans = -1;    cout << ans << endl;    return 0;}