hud上的动态规划练习

1、hud 2955 Robberies
（转载，你会发现dp下标是什么以及内容存储什么是很重要的）

第一点就是最容易犯错的一点，把所求总被抓概率看成背包容量。这就有两个很明显错误，求的就是1-被抓概率=逃跑概率，限制条件和最大背包容量不能同为一属性。再者，概率为多为浮点数，精确度不高无法遍历。所以背包容量必然是钱数，因为他能抢的银行有限，钱数也是有限的。然后是求最大逃跑概率，而题中每项给出的是被抓概率，所以要先被1减一下。还有最后求得的逃跑概率随着抢银行的数量增加而减少，多抢一个银行，其钱数必将转化为概率的乘积，所以动态方程也要做出改变。最后遍历，剩余的钱数越多，说明所抢的钱数越少，逃跑几率越大。所以从大到小遍历背包容量，一旦大于p，即为最大概率跳出。

#include <stdio.h>#include <algorithm>#include <iostream>#include <cmath>#include <string.h>using namespace std;const int N = 50005;int main(){  //  freopen("in.txt", "r", stdin);    int t, m0, m[N];    double p0, p[N], ans[N];    scanf("%d", &t);    while(t --)    {        scanf("%lf%d", &p0, &m0);        int sum = 0;        for(int i = 0; i < m0; i ++)        {            scanf("%d%lf", &m[i], &p[i]);            sum += m[i];        }        memset(ans, 0, sizeof(ans));        ans[0] = 1;        for(int i = 0; i < m0; i ++)        {            for(int j = sum; j >= m[i]; j --)            {                ans[j] = max(ans[j], ans[j - m[i]] *(1 - p[i]));            }        }        for(int i = sum; i >= 0; i --)        {            if(ans[i] > (1 - p0))            {                printf("%d\n", i);                break;            }        }    }    return 0;}

2、hud 1864 最大报销额
最主要的是将数据扩大100倍 dp[j]=max(dp[j],dp[j-v[i]]+v[i])

#include<iostream>#include<cstdio>#include<cstring>#include<vector>#include<queue>#include<algorithm>#define MAXNUM 3000050using namespace std;int N, dp[MAXNUM], bill[35];float sum;int main(){    //freopen("1.txt", "r", stdin);    int i, j, k, t, res, total, n[3];    char c;    float cash;    bool flag;    while (scanf("%f%d", &sum, &N) != EOF&&N)    {        memset(dp, 0, sizeof(dp));        memset(bill, 0, sizeof(bill));        total = (int)(100 * sum);        t = 0;        for (i = 0; i < N; i++)        {            memset(n, 0, sizeof(n));            flag = 1;            scanf("%d", &k);            for (j = 0; j < k; j++)            {                scanf(" %c:%f", &c, &cash);                if (c == 'A' || c == 'B' || c == 'C')                    n[c - 'A'] += (int)(cash * 100);                else                    flag = 0;            }            res = n[0] + n[1] + n[2];            if (n[0] <= 60000 && n[1] <= 60000 && n[2] <= 60000 && flag&&res<=100000)                bill[t++] = res;        }        for (i = 0; i < t; i++)        {            for (j = total; j >= bill[i]; j--)            {                dp[j] = max(dp[j - bill[i]] + bill[i], dp[j]);            }        }        printf("%.2f\n", float(float(dp[total]) / 100));    }    return 0;}

3、hdu 1231 最大连续子序列

#include<iostream>#include<cstdio>#include<cstring>#include<vector>#include<queue>#include<algorithm>#define MAXNUM 10010using namespace std;int N, dp[MAXNUM], start[MAXNUM], num[MAXNUM];int main(){    //freopen("1.txt", "r", stdin);    int i, j, k, t, ans, total, end;    bool flag;    while (scanf("%d", &N) && N)    {        memset(dp, 0, sizeof(dp));        memset(start, 0, sizeof(start));        memset(num, 0, sizeof(num));        flag = 1;        for (i = 1; i <= N; i++)        {            scanf("%d", &num[i]);            if (flag&&num[i] >= 0)                flag = 0;        }        if (flag)        {            printf("0 %d %d\n", num[1], num[N]);            continue;        }        ans = dp[1] = num[1];        start[1] = 1;        end = 1;        for (i = 2; i <= N; i++)        {            if (dp[i - 1] >= 0)            {                start[i] = start[i - 1];                dp[i] = num[i] + dp[i - 1];            }            else            {                start[i] = i;                dp[i] = num[i];            }            if (dp[i] > ans)            {                ans = dp[i];                end = i;            }        }        printf("%d %d %d\n", ans, num[start[end]], num[end]);    }    return 0;}

4、hdu 1506 Largest Rectangle in a Histogram

最开始我想的是每一个都去比较一下，结果超时了

for (i = 2; i <= N; i++)        {            dp[i] = tmp = num[i];            t = 1;            for (j = i - 1; j >= 1; j--)            {                tmp = min(tmp, num[j]);                dp[i] = max(dp[i], tmp*(++t));            }            ans = max(dp[i], ans);        }

结果看了网上的别人的代码才发现，dp[i]每次记录的是包含i矩形的最大值即可。
最主要的还是引进了L数组和R数组，类似于find_union的思想；

while (num[i] <= num[j + 1] &&j < N)                j = R[j + 1];

这个代码之前写错了，应该是j

#include<iostream>#include<cstdio>#include<cstring>#include<vector>#include<queue>#include<algorithm>#define MAXNUM 100010using namespace std;int N;long long num[MAXNUM], L[MAXNUM], R[MAXNUM];int main(){    freopen("1.txt", "r", stdin);    int i, j, k, t, l, r;    long long ans, tmp;    while (scanf("%d", &N)!=EOF && N)    {        for (i = 1; i <= N; i++)        {            scanf("%lld", &num[i]);        }           L[1] = 1;        R[N] = N;        for (i = 2; i <= N; i++)        {            j = i;            while (num[j - 1] >= num[i] && j > 1)                j = L[j - 1];            L[i] = j;        }        for (i = N-1; i >= 1; i--)        {            j = i;            while (num[i] <= num[j + 1] &&j < N)                j = R[j + 1];            R[i] = j;        }        ans = 0;        for (i = 1; i <= N; i++)        {            tmp = (R[i] - L[i] + 1)*num[i];            if (tmp > ans)                ans = tmp;        }        printf("%lld\n", ans);    }    return 0;}