HOJ1215 HDU1203

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两道题都是01背包的入门题。共同点：求放入背包物品的最小值。区别：前者的容量有下限（这和一般的背包相反）。

有一个tip：使用memset函数要注意，它并不是简单的初始化，慎用。

HOJ1215

转化为求”使得血量达到100所消耗分数的最小值“

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;const int INF = 0x3f3f3f3f;const int MAX = 10010;int dp[10020];int main(){    int t;    scanf("%d", &t);    while(t--){        int hp;        scanf("%d", &hp);        int n, sum = 0;        scanf("%d", &n);        int c[10020], v[10200];        for(int i = 1; i <= n; i++){            scanf("%d %d", &c[i], &v[i]);            sum += v[i];        }        for(int i = 1; i <= MAX; i++) dp[i] = INF;        for(int i = 1; i <= n; i++)            for(int j = MAX; j >= c[i]; j--)            dp[j] = min(dp[j], dp[j - c[i]] + v[i]);        int res = INF;        for(int i = hp; i <= MAX; i++) res = min(res, dp[i]);        if(res == INF) res = 0;        else res = sum - res;        printf("%d\n", res);    }    return 0;}

HDU1203

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;int main(){    int n, m;    while(scanf("%d %d", &n, &m) != EOF && (n || m)){        int c[10005];        double v[10005];        for(int i = 1; i <= m; i++){            scanf("%d %lf", &c[i], &v[i]);            v[i] = 1 - v[i];        }        double dp[10010];        for(int i = 0; i <= 10005; i++) dp[i] = 1;        for(int i = 1; i <= m; i++)            for(int j = n; j >= c[i]; j--)            dp[j] = min(dp[j], dp[j - c[i]] * v[i]);        printf("%.1lf", 100 * (1 - dp[n]));        puts("%");    }    return 0;}

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