#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;const int maxn = 1000 + 5;int n, m;
int value[maxn];
int dp[maxn];void input() {for(int i = 0; i < n; ++i) {cin >> value[i];}cin >> m;
}void solve() {if(m < 5) {cout << m << endl;return ;}m -= 5;sort(value, value + n);memset(dp, 0, sizeof(dp));for(int i = 0; i < n - 1; ++i) {for(int j = m; j >= value[i]; --j) {dp[j] = max(dp[j - value[i]] + value[i], dp[j]);}}cout << m + 5 - dp[m] - value[n - 1] << endl;
}int main() {while(cin >> n && n != 0) {input();solve();}return 0;
}

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 20096    Accepted Submission(s): 7443

Problem Description

The aspiring Roy the Robber has seen a lot of American movies, and knows that the bad guys usually gets caught in the end, often because they become too greedy. He has decided to work in the lucrative business of bank robbery only for a short while, before retiring to a comfortable job at a university.

For a few months now, Roy has been assessing the security of various banks and the amount of cash they hold. He wants to make a calculated risk, and grab as much money as possible.

His mother, Ola, has decided upon a tolerable probability of getting caught. She feels that he is safe enough if the banks he robs together give a probability less than this.

Input

The first line of input gives T, the number of cases. For each scenario, the first line of input gives a floating point number P, the probability Roy needs to be below, and an integer N, the number of banks he has plans for. Then follow N lines, where line j gives an integer Mj and a floating point number Pj . 
Bank j contains Mj millions, and the probability of getting caught from robbing it is Pj .

Output

For each test case, output a line with the maximum number of millions he can expect to get while the probability of getting caught is less than the limit set.

Notes and Constraints
0 < T <= 100
0.0 <= P <= 1.0
0 < N <= 100
0 < Mj <= 100
0.0 <= Pj <= 1.0
A bank goes bankrupt if it is robbed, and you may assume that all probabilities are independent as the police have very low funds.

Sample Input

3 0.04 3 1 0.02 2 0.03 3 0.05 0.06 3 2 0.03 2 0.03 3 0.05 0.10 3 1 0.03 2 0.02 3 0.05

Sample Output

2 4 6

题意：抢匪抢劫银行j的金额是Mj，被抓的概率是Pj。当被抓概率小于P时，认为抢匪是可以逃脱的。那么问题来了，在可逃脱的情况下，抢匪最多能抢多少钱。

分析：01背包。一眼看去是裸01背包，但是数组下标却是浮点数，所以得想办法转化。我们把银行的金额总和当做背包容量，不被抓的概率当做价值，那么转移方程就是：dp[j] = max(dp[j - p[i]] * (1 - m[i]), dp[j])。那么结果也就显而易见了，找到最大的容量V，dp[V] > 1 - P，那么这个V就是结果了。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2955

代码清单：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;#define EPS 0.000000001const int maxn = 100 + 5;int T;
double P;
int N;
int p[maxn];
double m[maxn];
double dp[maxn * maxn];void input() {cin >> P >> N;for(int i = 0; i < N; ++i) {cin >> p[i] >> m[i];}
}void solve() {if(P - 0 < EPS) {cout << "0" << endl;return ;}int V = 0;for(int i = 0; i < N; ++i) {V += p[i];}dp[0] = 1;for(int i = 1; i <= V; ++i) dp[i] = 0;for(int i = 0; i < N; ++i) {for(int j = V; j >= p[i]; --j) {dp[j] = max(dp[j - p[i]] * (1 - m[i]), dp[j]);}}for(int i = V; i >= 0; --i) { if(dp[i] - (1 - P) > EPS) {cout << i << endl;return ;}}
}int main() {cin >> T;for(int t = 0; t < T; ++t) {input();solve();}return 0;
}

HDU_1203 I NEED A OFFER!

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;const int maxn = 10000 + 5;int n, m;
int a[maxn];
double b[maxn];
double dp[maxn];void input() {for(int i = 0; i <m; ++i) {scanf("%d%lf", &a[i], &b[i]);}
}void solve() {dp[0] = 1;for(int i = 1; i <= n; ++i) dp[i] = 1;for(int i = 0; i < m; ++i) {for(int j = n; j >= a[i]; --j) {dp[j] = min(dp[j - a[i]] * (1 - b[i]), dp[j]);//cout << dp[j] << " ";}//cout << endl;}printf("%.1lf%%\n", (1 - dp[n]) * 100);
}int main() {while(scanf("%d%d", &n, &m) != EOF && n+m) {input();solve();}return 0;
}

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 35145    Accepted Submission(s): 12188

Problem Description

Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don't know that Computer College had ever been split into Computer College and Software College in 2002.
The splitting is absolutely a big event in HDU! At the same time, it is a trouble thing too. All facilities must go halves. First, all facilities are assessed, and two facilities are thought to be same if they have the same value. It is assumed that there is N (0<N<1000) kinds of facilities (different value, different kinds).

Input

Input contains multiple test cases. Each test case starts with a number N (0 < N <= 50 -- the total number of different facilities). The next N lines contain an integer V (0<V<=50 --value of facility) and an integer M (0<M<=100 --corresponding number of the facilities) each. You can assume that all V are different.
A test case starting with a negative integer terminates input and this test case is not to be processed.

Output

For each case, print one line containing two integers A and B which denote the value of Computer College and Software College will get respectively. A and B should be as equal as possible. At the same time, you should guarantee that A is not less than B.

Sample Input

2 10 1 20 1 3 10 1 20 2 30 1 -1

Sample Output

20 10 40 40

题意：把所有电脑分成A，B部分，使得A、B两部分金额尽可能相近，并保证金额 A > B。

分析：01背包。把总价平分一下，较小的一部分当做B背包的容量，那么对B进行01背包即可得到B的最大价值，A的价值也就出来了。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1171

代码清单：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;const int maxn = 1000 + 5;
const int maxm = 100 + 5;int N;
int V[maxn];
int M[maxn];
int value[maxn * maxm];
int dp[maxn * maxm];void input() {for(int i = 0; i < N; ++i) {scanf("%d%d", &V[i], &M[i]);}
}void solve() {int sumV = 0, va = 0;for(int i = 0; i < N; ++i) {sumV += V[i] * M[i];for(int j = 0; j < M[i]; ++j) {value[va++] = V[i];}}int minV = sumV / 2;memset(dp, 0, sizeof(dp));for(int i = 0; i < va; ++i) {for(int j = minV; j >= value[i]; --j) {dp[j] = max(dp[j - value[i]] + value[i], dp[j]);}}printf("%d %d\n", sumV - dp[minV], dp[minV]);
}int main() {while(scanf("%d", &N) != EOF && N >= 0) {input();solve();}return 0;
}

HDU_2639 Bone Collector II

代码清单：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;const int maxn = 100 + 5;
const int maxk = 30 + 5;
const int maxv = 1000 + 5;int T;
int N, V, K;
int value[maxn];
int volume[maxn];int dp[maxv][maxk];
int pre[maxk], now[maxk];void input() {cin >> N >> V >> K;for(int i = 0; i < N; ++i) {cin >> value[i];}for(int i = 0; i < N; ++i) {cin >> volume[i];}
}void solve() {memset(dp, 0, sizeof(dp));pre[K + 1] = now[K + 1] = -1;for(int i = 0; i < N; ++i) {for(int j = V; j >= volume[i]; --j) {for(int k = 1; k <= K; ++k) {pre[k] = dp[j][k];now[k] = dp[j - volume[i]][k] + value[i];}int idxp = 1, idxn = 1, idxk = 1;while(idxk <= K && (idxp <= K || idxn <= K)) {if(pre[idxp] >= now[idxn]) {dp[j][idxk] = pre[idxp++];}else {dp[j][idxk] = now[idxn++];}//Remove duplicate numberif(idxk == 1 || dp[j][idxk] != dp[j][idxk - 1]) { idxk += 1;}}}}cout << dp[V][K] << endl;
}int main() {cin >> T;for(int t = 0; t < T; ++t) {input();solve();}return 0;
}