Vijos P1836hys and Tanabata Festival big battle (01 backpack 2--convert DP object)

Title: Vijos P1836hys and Tanabata festival big battle

Test instructions

n objects, per value VI, specific gravity PI, total capacity 100

Analysis:

Similar to the weight of the Backpack pi is a real number, not as subscript, so change the DP object

Will find the maximum value in capacity 100 → to find the minimum capacity of the corresponding value,

Then the value of the first ≤100 of the volume, which is the highest value of the value that meets the criteria

Status: Dp[v]: Minimum volume with a value of V

Transfer equation:

Dp[v] = min (Dp[v], Dp[v-v[i]] + p[i]);

Core:

for (i = 1; i<=n; i++)
{
for (j = sum_v; j>=v[i]; j--)
{
Dp[j] = min (Dp[j], Dp[j-v[i]]    + p[i]);
}
}

Code:

#include  <stdio.h> #include  <iostream> #include  <math.h> #include  < algorithm> #include  <string.h> #include  <string> #include  <queue> #include  <stack> #include  <map> #include  <vector> #include  <time.h> using

 namespace std;
double p[1000+10];
int v[1000+10];

double dp[5*1000+10];

	Int main () {//freopen ("a.txt",  "R",  stdin);
	int n, i, j;
		while (~SCANF ("%d",  &n)) {int sum = 0;
			for (i = 1; i<=n; i++) {scanf ("%lf%d",  &p[i], &v[i]);
		sum += v[i];
		} memset (Dp, 0x4f, sizeof (DP));
		dp[0] = 0; for (i = 1; i<=n; i++) {for (j = sum; j>=v[i]; j--) {dp[j] 
			= min (Dp[j], dp[j-v[i]] + p[i]);
}} while (dp[sum]>100) sum--;		printf ("%d\n",  sum);
} return 0; }