0-1背包

关于HDU 2602 ，按照算法课上讲的写了一下

二维：

#include <stdio.h>
#include <string.h>

int n,c;

int v[1001];
int w[1001];
int dp[1001][1001];

int max (int a, int b) {
	return (a>b)? a:b;
}

int min(int a, int b) {
	return (a>b)? b:a;
}

void knapsack() {
	memset(dp,0,sizeof(dp));
	int jMax;
	for (int i=0; i<=c; ++i) dp[0][i] = 0;
	for (int i=1; i<n; ++i) {
		jMax = min(w[i]-1, c);
		for (int j=0; j<=jMax; ++j) {
			dp[i][j] = dp[i-1][j];
		}
		for (int j=w[i]; j<=c; ++j) {
			dp[i][j] = max(dp[i-1][j], dp[i-1][j-w[i]] + v[i]);
		}
	}
	dp[n][c] = dp[n-1][c];
	if(c>=w[n]) dp[n][c] = max(dp[n][c], dp[n-1][c-w[n]] + v[n]);
}

int main () {
	int ncase;
	scanf("%d",&ncase);
	while (ncase--) {
		scanf("%d%d", &n, &c);
		for (int i=1; i<=n; ++i) scanf("%d",v+i);
		for (int i=1; i<=n; ++i) scanf("%d",w+i);
		knapsack();
		printf("%d\n",dp[n][c]);
		//for (int i=0; i<=n; ++i) {
		//	for (int j=0; j<=c; ++j) {
		//		printf("%4d ",dp[i][j]);
		//	}
		//	putchar('\n');
		//}
	}
}

其实对于0-1背包。。一维完全可以解决问题：

#include <stdio.h>
#include <string.h>

int n,c;

int v[1001];
int w[1001];
int dp[1001];

int max(int a, int b) {return a>b? a:b;}
int main () {
	int ncase;
	scanf("%d",&ncase);
	while (ncase--) {
		memset(dp,0,sizeof(dp));
		scanf("%d%d", &n, &c);
		for (int i=1; i<=n; ++i) scanf("%d",v+i);
		for (int i=1; i<=n; ++i) scanf("%d",w+i);
		for (int i=1; i<=n; ++i)
			for (int j=c; j>=w[i]; --j) dp[j] = max(dp[j],dp[j-w[i]] + v[i]);
		printf("%d\n",dp[c]);
	}
}

典型的  0-1背包问题  还有HDU 1203

#include <stdio.h>
#include <string.h>

int n,m;

int w[10001];
double v[10001];
double dp[10001];

double min(double a, double b) {
	return a>b? b:a;
}

int main (){
	while (scanf("%d%d",&n,&m),n+m) {
		for (int i=0; i<=n; ++i) dp[i] = 1.0;
		for (int i=1; i<=m; ++i)
			scanf("%d%lf", &w[i], &v[i]);
		for (int i=1; i<=m; ++i) {
			for (int j=n; j>=w[i]; --j) {
				dp[j] = min(dp[j], dp[j-w[i]]*(1.0-v[i]));
			}
		}
		dp[n] = 1.0 - dp[n];
		dp[n] *= 100.0;
		printf("%.1lf%%\n",dp[n]);
	}
}

这题要求至少得到一份的最大概率， 那我们求得不到offer的最小概率即可~  如代码，01背包 dp存最小概率 ， 最后用1减，即可