UVa 242 Stamps and Envelope Size (无限背包,DP)

时间:2022-07-19 07:29:32

题意:信封上最多贴S张邮票。有N个邮票集合,每个集合有不同的面值。问哪个集合的最大连续邮资最 大,输出最大连续邮资和集合元素。

最大连续邮资是用S张以内邮票面值凑1,2,3...到n+1凑不出来了,最大连续邮资就是n。如果不止一个集合结果相 同,输出集合元素少的,

如果仍相同,输出最大面值小的。

析:这个题,紫书上写的不全,而且错了好几次,结果WA好几次。

首先这个和背包问题差不多,我们只用一维就好。dp[i]表示邮资为 i 时的最小邮票数,然后,如果dp[i] > s就该结束了。

其他的就很简单了,主要是我没理解题意。

代码如下:

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std; typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 15 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
int a[maxn][maxn];
int dp[1024];
int ans[maxn]; int main(){
while(scanf("%d", &n) == 1 && n){
scanf("%d", &m);
for(int i = 0; i < m; ++i){
scanf("%d", &a[i][0]);
for(int j = 1; j <= a[i][0]; ++j)
scanf("%d", &a[i][j]);
fill(dp, dp+1024, INF);
dp[0] = 0;
for(int j = 1; j < 1024; ++j){
for(int k = 1; k <= a[i][0] && j-a[i][k] >= 0; ++k){
dp[j] = Min(dp[j], dp[j-a[i][k]]+1);
}
if(dp[j] > n){ ans[i] = j-1; break; }
}
} int anss = -1, cnt = 0;
for(int i = 0; i < m; ++i){
if(ans[i] > anss){ anss = ans[i]; cnt = i; }
else if(ans[i] == anss && a[i][0] < a[cnt][0]) cnt = i;
else if(ans[i] == anss && a[i][0] == a[cnt][0]){
bool ok = false;
for(int j = a[i][0]; j >= 0; --j)
if(a[i][j] < a[cnt][j]){ ok = true; break; }
else if(a[i][j] > a[cnt][j]) break;
if(ok) cnt = i;
}
}
printf("max coverage =%4d :", anss);
for(int i = 1; i <= a[cnt][0]; ++i) printf("%3d", a[cnt][i]);
printf("\n");
}
return 0;
}