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

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;}

 

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