LightOJ 1079 Just another Robbery 概率dp

/*    题目描述：有n件物品，盗走第i件物品可以得到财产m[i]，但同时有p[i]的概率被抓住，现要求从这n件物品中取走若干件，             使得得到的财产量最多，又要保证被抓住的概率小于P    思路：dp[i][j]表示到第i件物品时得到财产j时被抓住的概率最小值，则有         dp[i][j] = min(dp[i - 1][j] , dp[i - 1][j - m[i]] + (1 - dp[i][ j - m[i] ] ) * p[i])         min中第一项含义为从i - 1件中已经拿到了财产j，那么第i件不必冒风险去拿，所以dp[i][j] = dp[i - 1][j - 1]，         第二项含义为从i-1 件中已经拿到了财产j - m[i] ， 再拿走第i件就可以补全财产总量j，那么被抓到的概率等于在拿到j - m[i]         过程中被抓到的概率 + 在拿j - m[i]过程中未被抓到的概率 * 拿第i件物品时被抓到的概率*/#pragma warning(disable:4786)#pragma comment(linker, "/STACK:102400000,102400000")#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<stack>#include<queue>#include<map>#include<set>#include<vector>#include<cmath>#include<string>#include<sstream>#define LL long long#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)#define mem(a,x) memset(a,x,sizeof(a))#define lson l,m,x<<1#define rson m+1,r,x<<1|1using namespace std;const int INF = 0x3f3f3f3f;const int mod = 1e9 + 7;const double PI = acos(-1.0);const double eps=1e-8;const int maxn = 1e4 + 5;double dp[maxn] , p[maxn];int m[maxn];int main(){    int n , T , kase = 0;    double P;    scanf("%d",&T);    while(T--){        scanf("%lf%d",&P , &n);        int sum = 0;        for(int i = 1 ; i <= n ; i++){            scanf("%d%lf",&m[i],&p[i]);            sum += m[i];        }        for(int i = 0 ; i<= sum ; i++)            dp[i] = INF;        dp[0] = 0;        sum = 0;        int ans = - 1;        for(int i = 1 ; i <= n ; i++ ){            sum += m[i] ;            for(int j = sum ; j >= 1 ; j--){                if(j >= m[i] && (0 <= dp[j - m[i]]&&dp[j - m[i] ]<= 1))                    dp[j] = min(dp[j - m[i]] + (1.0 - dp[j - m[i]] ) * p[i] , dp[j]);                if(dp[j] < P)       ans = max(ans , j);            }        }        if(ans != -1)            printf("Case %d: %d\n" , ++kase , ans );        else            printf("Case %d: 0\n" , ++kase );    }    return 0;}