【背包】POJ 1882

題目有點囉嗦，說白了給出一個上限s，n套郵票，每套有m張面值為ai的郵票，問不超過上限的前提下能組成最大的連續面值和（從1開始）。

dp[N]：組成面值為i的最小郵票數量+完全背包，更新幾個記錄值

#define N 1100int dp[N];//組成面值為i的最小數量int a[12][12];int main(){    int s;    while(scanf("%d",&s) && s){        int i,j,k;        int n;        scanf("%d",&n);        int ans = MIN;        int index = 0;        int num = MAX;        int big = 0;        for(i=1;i<=n;i++){            scanf("%d",&a[i][0]);            int m = a[i][0];            for(j=1;j<=m;j++){                scanf("%d",&a[i][j]);            }            dp[0] = 0;            int maxv = a[i][m]*s;            for(j=1;j<=maxv;j++)dp[j] = (1<<30);            for(j=1;j<=m;j++){                for(k=a[i][j];k<=maxv;k++){                    dp[k] = min(dp[k],dp[k-a[i][j]] + 1);                }            }            for(j=1;j<=maxv;j++){                if(dp[j]>s)break;            }            j--;            if(j>ans){                ans = j;                index = i;                num = m;                big = a[i][m];            } else if(j == ans){                if(m<num){                    ans = j;                    index = i;                    num = m;                    big = a[i][m];                } else if(m == num){                    if(a[i][m]<big){                        ans = j;                        index = i;                        num = m;                        big = a[i][m];                    }                }            }        }        printf("max coverage = %d :",ans);        for(i=1;i<=num;i++){            printf(" %d",a[index][i]);        }        puts("");    }    return 0;}