首页 > 代码库 > PKU 2184 Cow Exhibition 01背包

PKU 2184 Cow Exhibition 01背包

题意:

  有一些牛,每头牛有一个Si值,一个Fi值,选出一些牛,使得max( sum(Si+Fi) ) 并且 sum(Si)>=0, sum(Fi)>=0

思路:

  随便选一维做容量(比如Fi),另一维做价值,然后直接做01背包。

  做的时候注意一下方向。

  最后,在合法解里面找一下最优解就好了。

代码:

  

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <algorithm>#include <string>#include <queue>#include <stack>#include <vector>#include <map>#include <set>#include <functional>#include <cctype>#include <time.h>using namespace std;typedef pair<int, int> PII;const int INF = 1e6;const int MAXM = 1e5+5;const int MAXN = 111;PII a[MAXN];int dp[MAXM];int n;int main() {    #ifdef Phantom01        freopen("PKU2184.txt", "r", stdin);    #endif //Phantom01    while (scanf("%d", &n)!=EOF) {        int sum = 0;        for (int i = 0; i < n; i++) {            scanf("%d%d", &a[i].first, &a[i].second);            if (a[i].first>0) sum += a[i].first;        }        sort(a, a+n);        for (int i = 0; i <= sum; i++) dp[i] = -INF;        dp[0] = 0;        for (int i = n-1; i >= 0; i--) {            int f = a[i].first, s = a[i].second;            if (f<0 && s<0) continue;            if (f>=0) {                for (int j = sum-f; j >= 0; j--) {                    dp[j+f] = max(dp[j+f], dp[j]+s);                }            } else {                for (int j = -f; j <= sum; j++) {                    dp[j+f] = max(dp[j+f], dp[j]+s);                }            }        }        int ans = 0;        for (int i = 0; i <= sum; i++) if (dp[i]>=0) {            ans = max(ans, i+dp[i]);        }        printf("%d\n", ans);    }    return 0;}
View Code

 

写的时候又一次为了剪枝,不小心删掉了一些合法解,甚至是最优解,然后wa了两发……真是伤心

PKU 2184 Cow Exhibition 01背包