动态规划 hdu2955 思考过程

看下文之前最好先阅读 <算法导论> 第15章, <背包问题九讲>第一讲 (推荐)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2955
根据题目的已知条件, 该问题可以使用基本01背包模型
    银行数量:N
    每家银行所拥有的金钱数量: mounts[i]
    盗取各家银行的风险指数: risks[i]
    个人所能承担的风险的极限: V
    求在风险范围内, 盗取最多的资金
    
但该问题有两个问题待解决:
    1.风险极限V是个浮点数; 一方面精度无法得知, 另一方面, 编程语言在处理浮点数时存在一定的难度
    2.题目中提到"the probability of getting caught from robbing", 是指被抓的概率. 设盗取第i家银行的概率是Pi, 盗取第i家银行逃跑的概率是Ri
    则 Pi = 1 - Ri
    且盗取多家(i家)银行被抓的概率 PPi = 1 - (R1*R2*R3*...*Ri), 这中间有个转换过程
    
    对于第一个问题, 采用的解决方案是将多家银行的总资金数量作为01背包问题中的背包重量V, 可以这样转换的原因是 
    转换之前, 我们求解的是 资金关于风险的离散函数即y=money[x], 且是单调的; 
    转换之后, 我们求解的是 风险关于资金的离散函数即x=money'[y], 并且也是单调的.

//    源码:

#include <stdio.h>
#include <stdlib.h>
#include <memory.h>

int main()
{
    int ncases = 0;
    int nbanks = 0;
    int *mounts;
    double deadline = 0.0;
    double *risks;

    /*	process the input  */
    setbuf(stdin, NULL);
    while ( scanf("%d", &ncases) != EOF ) {
        if ( ncases < 1 || ncases > 100 ) {
            printf("Invalid ncases\n");
            setbuf(stdin, NULL);
            continue;
        }
        while( ncases ) {
        scanf("%lf %d", &deadline, &nbanks);
        if ( deadline < 0.0 || deadline > 1.0 
        || nbanks < 1 || nbanks > 100 ) {
            printf("Invalid deadline or nbanks\n");
            setbuf(stdin, NULL);
            ncases--;
            continue;
        }
        
        mounts = (int *)malloc(nbanks*sizeof(int));
        risks = (double *)malloc(nbanks*sizeof(double));
        int i, total = 0;
        for ( i = 0; i < nbanks; i++ ) {
            scanf("%d %lf", &mounts[i], &risks[i]);
            if ( mounts[i] < 1 || mounts[i] > 100 
            || risks[i] < 0.0 || risks[i] > 1.0 ) {
                printf("Invalid mounts or risks\n");
                setbuf(stdin, NULL);
                free(mounts);
                free(risks);
                ncases--;
                continue;
            }
            total += mounts[i];
        }

	/*	core algorithm	*/

	/*
	**	runaway probality
	**	the index means the mounts robbed
	**	runaway[0] means that the robber doesn't rob bank
	**	runaway[v] = max{runaway[v], runaway[v-mounts[i]]*risks[i]}
	*/
        double *runaway = (double *)malloc((total+1)*sizeof(double));
        runaway[0] = 1.0;
        for ( i = 1; i <= total; i++ )
            runaway[i] = 0.0;

        int j;
        double tmp = 0.0;
        for ( i = 0; i < nbanks; i++ ) {
            for ( j = total; j >= 0; j-- ) {
                if ( j >= mounts[i] )
                    tmp = runaway[j-mounts[i]]*(1-risks[i]);
                else
                    tmp = runaway[j];
                runaway[j] = tmp > runaway[j] ? tmp : runaway[j];
            }
        }

        int k;
        for ( k = total; k >= 0; k-- )
            if ( runaway[k] >= 1.0-deadline ) {
                printf("%d\n", k);
                break;
            }
	/*  clear all */
        free(mounts);
        free(risks);
        free(runaway);
        ncases--;
    } 
    }
    
    return 0;
}


    参考资料:
    1.http://blog.youkuaiyun.com/qq172108805/article/details/8141851