LightOJ - 1079 Just another Robbery(背包)

题目大意：给出一个概率P，和N个银行，再给出每个银行的被捕概率和钱的数量，问在被捕概率小于P的情况下，偷到的最多的钱

解题思路：用dp[i][j]表示前i个银行，偷盗了j的钱的最低被捕概率，接着进行背包

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 110;const int M = 10010;const double esp = 1e-6;double p;int n, sum, cas = 1;int money[N];double caught[N], dp[N][M];void init() {    scanf("%lf%d", &p, &n);    sum = 0;    for (int i = 1; i <= n; i++) {        scanf("%d%lf", &money[i], &caught[i]);        sum += money[i];    }}void solve() {    memset(dp, -1, sizeof(dp));    dp[0][0] = 0;    for (int i = 1; i <= n; i++)        for (int j = 0; j <= sum; j++) {            //第i个银行不抢            dp[i][j] = dp[i - 1][j];            //抢            if (j - money[i] >= 0 && (double)1 + dp[i - 1][j - money[i]] > esp) {                if ((double)1 + dp[i][j] > esp)                     dp[i][j] = min(dp[i][j], dp[i - 1][j - money[i]] + (1 - dp[i - 1][j - money[i]]) * caught[i]);                else                     dp[i][j] = dp[i - 1][j - money[i]] + (1 - dp[i - 1][j - money[i]]) * caught[i];            }        }    for (int i = sum; i >= 0; i--)        if (p - dp[n][i] > esp) {            printf("Case %d: %d\n", cas++, i);            break;        }}int main() {    int test;    scanf("%d", &test);    while (test--) {        init();        solve();    }    return 0;}