poj1015 Jury Compromise

dp题，类似于01背包的转移
需要注意的是：背包容量在有的时候可能为负数，所以需要算出最大数据量整体平移
像01背包一样直接倒序循环j并不能保证每一个物品只选一个，两个等价的物品在计算时可能会重复使用
所以把i（物品）放在最后一维，通过chk函数来检查该物品此前是否已经选过，从而保证转移的正确性
继续锻炼思维……

Code：

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

const int N = 205;
const int M = 25;
const int MaxNum = 805;

int testcase = 0, n, m, a[N], b[N], dp[M][MaxNum], path[M][MaxNum], tot, s[M];

inline void read(int &X) {
    X = 0;
    char ch = 0;
    int op = 1;
    for(; ch > '9' || ch < '0'; ch = getchar())
        if(ch == '-') op = -1;
    for(; ch >= '0' && ch <= '9'; ch = getchar())
        X = (X << 3) + (X << 1) + ch - 48;
    X *= op;
} 

void find(int j, int k) {
    if(j == 0) return;
    s[++tot] = path[j][k];
    find(j - 1, k - (a[path[j][k]] - b[path[j][k]]));
}

bool chk(int j, int k, int i) {
    if(j == 0) return 1;
    if(path[j][k] == i) return 0;
    return chk(j - 1, k - (a[path[j][k]] - b[path[j][k]]), i);
}

int main() {
    for(read(n), read(m); n + m != 0; read(n), read(m)) {
        testcase++;
        for(int i = 1; i <= n; i++)
            read(a[i]), read(b[i]);
            
        int fix = m * 20;
        memset(dp, 0xcf, sizeof(dp));
        memset(path, 0, sizeof(path));
        const int inf = -dp[0][0];
        dp[0][fix] = 0;
        
/*        for(int i = 1; i <= n; i++)
            for(int j = m; j >= 1; j--)    
                for(int k = 0; k <= 800; k++)     
                    if(dp[j - 1][k - (a[i] - b[i])] != -inf) 
                        if(chk(j - 1, k - (a[i] - b[i]), i))
                        if(dp[j - 1][k - (a[i] - b[i])] + (a[i] + b[i]) > dp[j][k]) {
                            dp[j][k] = dp[j - 1][k - (a[i] - b[i])] + (a[i] + b[i]);
                            path[j][k] = i;
                        } */
            
        for(int j = 1; j <= m; j++)
            for(int k = 0; k <= 2 * fix; k++)
                if(dp[j - 1][k] >= 0) 
                    for(int i = 1; i <= n; i++)
                        if(chk(j - 1, k, i) && dp[j - 1][k] + a[i] + b[i] > dp[j][k + (a[i] - b[i])]) {
                            dp[j][k + (a[i] - b[i])] = dp[j - 1][k] + a[i] + b[i];
                            path[j][k + (a[i] - b[i])] = i;
                        }        
        
        int ans;
        for(int i = 0; i <= fix; i++) {
            if(dp[m][fix + i] != -inf && dp[m][fix - i] == -inf) {
                ans = i;
                break;
            } else if(dp[m][fix - i] != -inf && dp[m][fix + i] == -inf) {
                ans = -i;
                break;
            } else if(dp[m][fix + i] != -inf && dp[m][fix - i] != -inf){
                if(dp[m][fix - i] > dp[m][fix + i]) ans = -i;
                else ans = i;
                break;
            }
        } 
        
        tot = 0;
        find(m, fix + ans);
        sort(s + 1, s + m + 1);
        int ansp = 0, ansd = 0;
        for(int i = 1; i <= m; i++) {
            ansp += a[s[i]];
            ansd += b[s[i]];
        }
        
        printf("Jury #%d\n", testcase);
        printf("Best jury has value %d for prosecution and value %d for defence: \n", ansp, ansd);
        for(int i = 1; i <= m; i++)
            printf(" %d", s[i]);
        puts("\n");
    } 
    return 0;

吐槽：sb输出格式，defence：后面的空格坑死我了