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
动态规划 hdu2955 思考过程
最新推荐文章于 2020-07-04 15:34:59 发布