POJ 2184 Cow Exhibition (01背包)

题意：每行给出si和fi，代表牛的两个属性，然后要求选出几头牛，是的则求出总S与总F的和，注意S与F都不能为负数

析：用dp[i]来表示存放s[i]的时最大的f[i]，其实就是一个01背包。只是取不取的关系。注意是有负数，所以把数组开大一点，然后s[i]的正负数，

我们取的顺序不同，正数是逆向，负数是正向，要不然可能有重复。还不知道为什么交G++就RE，交C++就能过。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<double, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 10;
const int mod = 1e6;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
int dp[maxn*2];
int s[105], t[105];

int main(){
while(scanf("%d", &n) == 1){
for(int i = 0; i < n; ++i)  scanf("%d %d", s+i, t+i);
memset(dp, -INF, sizeof dp);
dp[100000] = 0;
for(int i = 0; i < n; ++i){
if(s[i] <= 0 && t[i] <= 0)  continue;
if(s[i] > 0){
for(int j = 200000; j >= s[i]; --j)
if(dp[j-s[i]] > -INF)  dp[j] = max(dp[j], dp[j-s[i]]+t[i]);
}
else {
for(int j = s[i]; j <= 200000+s[i]; ++j)
if(dp[j-s[i]] > -INF)  dp[j] = max(dp[j], dp[j-s[i]]+t[i]);
}
}
int ans = 0;
for(int i = 100000; i <= 200000; ++i)
if(dp[i] >= 0)  ans = max(ans, dp[i] + i);
printf("%d\n", ans - 100000);
}
return 0;
}