hdu 4381（背包问题）

题意：有一排箱子全是黑色的，然后有两种操作1 a x从1-a跳出x个变成白色。2 a x从a-x跳出x个变成白色。要你求最多能变成白色的个数和最少的操作数。

思路：转化成背包问题，把两种操作分开进行背包，最后两重循环枚举找最优解即可。

代码如下：

/**************************************************
 * Author     : xiaohao Z
 * Blog     : http://www.cnblogs.com/shu-xiaohao/
 * Last modified : 2014-03-29 15:45
 * Filename     : MulSchool.cpp
 * Description     : 
 * ************************************************/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#define MP(a, b) make_pair(a, b)
#define PB(a) push_back(a)

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<unsigned int,unsigned int> puu;
typedef pair<int, double> pid;
typedef pair<ll, int> pli;
typedef pair<int, ll> pil;

const int INF = 0x3f3f3f3f;
const double eps = 1E-6;
const int LEN = 100000+10;
int T, n, m, ta, tb, dp1[LEN], dp2[LEN];
struct OP{
    int pos, x;
};
OP op1[LEN], op2[LEN];

bool cmp1(OP a, OP b){
    if(a.pos != b.pos) return a.pos < b.pos;
    else a.x < b.x;
}

bool cmp2(OP a, OP b){
    if(a.pos != b.pos) return a.pos > b.pos;
    else a.x < b.x;
}

int main()
{
//    freopen("in.txt", "r", stdin);

    int kase = 1;
    scanf("%d", &T);
    while(T--){
        scanf("%d%d", &n, &m);
        ta = tb = 0;
        memset(dp1, 0x3f, sizeof dp1);
        memset(dp2, 0x3f, sizeof dp2);
        for(int i=0; i<m; i++){
            int op = 0, a, b;
            scanf("%d%d%d", &op, &a, &b);
            if(op == 1) {
                op1[ta].pos = a;
                op1[ta].x = b;
                ta++;
            }else{
                op2[tb].pos = a;
                op2[tb].x = b;
                tb++;
            }
        }
        sort(op1, op1+ta, cmp1);
        sort(op2, op2+tb, cmp2);
        dp1[0] = dp2[n+1] = 0;
        for(int j=0; j<ta; j++){
            for(int i=n; i>=0; i--){
                if(op1[j].pos >= i && i - op1[j].x >= 0 && dp1[i - op1[j].x] != INF){
                    dp1[i] = min(dp1[i], dp1[i - op1[j].x]+1);
                }
            }
        }
        for(int j=0; j<tb; j++){
            for(int i=1; i<=n+1; i++){
                if(op2[j].pos <= i && i + op2[j].x <= n+1 && dp2[i + op2[j].x] != INF){
                    dp2[i] = min(dp2[i], dp2[i + op2[j].x]+1);
                }     
            }
        }
        int ansa = INF, ansb = INF;
        for(int i=0; i<=n; i++){
            for(int j=i+1; j<=n+1; j++){
                if(dp1[i] != INF && dp2[j] != INF){
                    if(j-i-1 < ansa){
                        ansa = j-i-1;
                        ansb = dp1[i] + dp2[j];
                    }
                    if(j-i-1 == ansa){
                        ansb = min(dp1[i] + dp2[j], ansb);
                    }
                }
            }
        }
        if(ansb == INF) ansb = 0;
        printf("Case %d: %d %d
", kase++, n-ansa, ansb);
    }
    return 0;
}