uva 242 DP 求s张邮票可以连续组合成最大面值多少的邮票

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int d[36][36],dp[1036],res[36];
int main(){
	int s,n;
	while(scanf("%d",&s) == 1 && s){
		scanf("%d",&n);
		for(int i = 1;i <= n;i++){
			scanf("%d",&d[i][0]);
			for(int j = 1;j <= d[i][0];j++)
			scanf("%d",&d[i][j]);
			memset(dp,0x3f,sizeof(dp));
			dp[0]=0;
			for(int j = 1;j <= 1001;j++){
				for(int k=1; k<=d[i][0];k++){
					if(j - d[i][k] >= 0)
					dp[j] = min(dp[j],dp[j-d[i][k]] + 1);
				}
				if(dp[j] > s){
					res[i] = j - 1;
					break;
				}
			}
		}
		int lo,num=-1;
		for(int i = 1;i <= n;i++){
			if(res[i] > num){
				lo = i;
				num = res[i];
			}
			else if(res[i] == num){
				if(d[i][0] < d[lo][0]){
					lo = i;
				}
				else if(d[i][0] == d[lo][0]){
					int flag=0;
					for(int j = d[i][0];j > 0;j--){
						if(d[i][j] != d[lo][j]){
							flag = d[i][j] < d[lo][j];
							break;
						}
					}
					if(flag) lo = i;
				}
			}
		}
		printf("max coverage = %4d :",num);
		for(int i = 1; i <= d[lo][0];i++)
		printf("%3d", d[lo][i]);
		printf("\n");
	}
}

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转载自blog.csdn.net/sky_zdk/article/details/78454276
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