POJ 1742 - Coins （ｄｐ 多重背包）

Coins

Time Limit: 3000MS Memory Limit: 30000KTotal Submissions: 28056 Accepted: 9492

Description

People in Silverland use coins.They have coins of value A1,A2,A3...An Silverland dollar.One day Tony opened his money-box and found there were some coins.He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact price(without change) and he known the price would not more than m.But he didn't know the exact price of the watch.
You are to write a program which reads n,m,A1,A2,A3...An and C1,C2,C3...Cn corresponding to the number of Tony's coins of value A1,A2,A3...An then calculate how many prices(form 1 to m) Tony can pay use these coins.

Input

The input contains several test cases. The first line of each test case contains two integers n(1<=n<=100),m(m<=100000).The second line contains 2n integers, denoting A1,A2,A3...An,C1,C2,C3...Cn (1<=Ai<=100000,1<=Ci<=1000). The last test case is followed by two zeros.

Output

For each test case output the answer on a single line.

Sample Input

3 101 2 4 2 1 12 51 4 2 10 0

Sample Output

84

Source

LouTiancheng@POJ

题意：

给出一些硬币的价值和数量，问用这些硬币能凑出多少种小于ｍ的面值

多重背包

两种方法

第一种比较简单好想容易理解

#include <cstdio>#include <iostream>#include <vector>#include <algorithm>#include <cstring>#include <string>#include <map>#include <cmath>#include <queue>#include <set>using namespace std;//#define WIN#ifdef WINtypedef __int64 LL;#define iform "%I64d"#define oform "%I64d\n"#elsetypedef long long LL;#define iform "%lld"#define oform "%lld\n"#endif#define S64I(a) scanf(iform, &(a))#define P64I(a) printf(oform, (a))#define REP(i, n) for(int (i)=0; (i)<n; (i)++)#define REP1(i, n) for(int (i)=1; (i)<=(n); (i)++)#define FOR(i, s, t) for(int (i)=(s); (i)<=(t); (i)++)const int INF = 0x3f3f3f3f;const double eps = 10e-9;const double PI = (4.0*atan(1.0));const int maxn = 100 + 20;const int maxm = 100000 + 20;int v[maxn];int c[maxn];bool vis[maxm];int use[maxm];int main() {    int n, m;    while(scanf("%d%d", &n, &m) != EOF && !(!n && !m)) {        for(int i=0; i<n; i++) scanf("%d", &v[i]);        for(int i=0; i<n; i++) scanf("%d", &c[i]);        memset(vis, 0, sizeof(vis));        int ans = 0;        vis[0] = true;        for(int i=0; i<n; i++) {            memset(use, 0, sizeof(use));            for(int j=v[i]; j<=m; j++) {                if(!vis[j] && vis[j-v[i]] && use[j-v[i]]<c[i]) {                    vis[j] = true;                    use[j] = use[j-v[i]] + 1;                    ans++;                }            }        }        printf("%d\n", ans);    }    return 0;}

另外一种

dp[i][j] 表示前ｉ种硬币凑出ｊ第ｉ种硬币剩余的数量

#include <cstdio>#include <iostream>#include <vector>#include <algorithm>#include <cstring>#include <string>#include <map>#include <cmath>#include <queue>#include <set>using namespace std;//#define WIN#ifdef WINtypedef __int64 LL;#define iform "%I64d"#define oform "%I64d\n"#elsetypedef long long LL;#define iform "%lld"#define oform "%lld\n"#endif#define S64I(a) scanf(iform, &(a))#define P64I(a) printf(oform, (a))#define REP(i, n) for(int (i)=0; (i)<n; (i)++)#define REP1(i, n) for(int (i)=1; (i)<=(n); (i)++)#define FOR(i, s, t) for(int (i)=(s); (i)<=(t); (i)++)const int INF = 0x3f3f3f3f;const double eps = 10e-9;const double PI = (4.0*atan(1.0));const int maxn = 100 + 20;const int maxm = 100000 + 20;int v[maxn];int c[maxn];int dp[maxm];int main() {    int n, m;    while(scanf("%d%d", &n, &m) != EOF && !(!n && !m)) {        for(int i=0; i<n; i++) scanf("%d", &v[i]);        for(int i=0; i<n; i++) scanf("%d", &c[i]);        memset(dp, -1, sizeof(dp));        dp[0] = 0;        for(int i=0; i<n; i++) {            for(int j=0; j<=m; j++) {                if(dp[j] >= 0) {                    dp[j] = c[i];                } else if(j < v[i] || dp[j-v[i]] <= 0) {                    dp[j] = -1;                } else {                    dp[j] = dp[j-v[i]] - 1;                }            }        }        int ans = 0;        for(int i=1; i<=m; i++) if(dp[i] >= 0) ans++;        printf("%d\n", ans);    }    return 0;}