01背包的路径全纪录

用DP求出 01背包的最大价值后该如何求得这条路径，就是有多少种选取情况下可以达到这个最大价值呢？

在本次百度之星的比赛中，有毒的1004让我好好思考了这个问题：

 

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;

int dp[105][1005];
int score[105], cost[105];
bool re[105][105];

int main()
{
    int CASE;
    scanf("%d", &CASE);
    for(int cas = 1; cas <= CASE; ++cas)
    {
        int pre, n;
        scanf("%d%d", &pre, &n);
        for(int i = 1; i <= n; ++i)
            scanf("%d%d", &score[i], &cost[i]);
        memset(dp, 0, sizeof(dp));
        memset(re, 0, sizeof(re));
        for(int i = 1; i <= n; ++i)
        {

            for(int j = pre; j >= cost[i]; --j)
            {
                if(dp[i - 1][j] <= dp[i - 1][j - cost[i]] + score[i])
                {
                    re[i][j] = 1;
                    dp[i][j] = dp[i - 1][j - cost[i]] + score[i];
                }
                else dp[i][j] = dp[i - 1][j];
            }
            for(int j = cost[i] - 1; j >= 0; --j)
                dp[i][j] = dp[i - 1][j];
        }
        int ans = dp[n][pre];

        int sum = 10005;
        int sum_cnt = 0;
        int con[105];

        for(int i = n; i >= 1; --i)
            for(int j = pre; j >= 0; --j)
            {
                if(dp[i][j] == ans && re[i][j])
                {
                    int tsum = 0;
                    int cnt = 0;
                    int tmp[105];
                    int x, y;
                    for(x = i, y = j; x >= 1 && y > 0; x -= 1)
                    {
                        if(re[x][y])
                        {
                            y -= cost[x];
                            tsum += x;
                            tmp[cnt++] = x;
                        }
                    }
                    if(y > 0) continue;
                    //for(int t = cnt - 1; t >= 0; --t)
                        //printf("%d%c", tmp[t], t == 0 ? '\n' : ' ');
                    if(tsum < sum)
                    {
                        sum = tsum;
                        sum_cnt = cnt;
                        for(int t = 0; t < cnt; ++t)
                            con[t] = tmp[t];
                    }
                    else if(tsum == sum)
                    {
                        int ok = 0;
                        for(int k = cnt - 1, t = sum_cnt - 1; k >= 0 && t >= 0; --t, --k)
                        {
                            if(con[t] != tmp[k])
                            {
                                ok = con[t] < tmp[k] ? 0 : 1;
                                break;
                            }
                        }
                        if(ok == 1)
                        {
                            sum_cnt = cnt;
                            for(int t = 0; t < cnt; ++t)
                                con[t] = tmp[t];
                        }
                    }
                }
            }
        sum = 0;
        for(int i = 0; i < sum_cnt; ++i) sum += cost[con[i]];
        printf("Case #%d:\n", cas);
        printf("%d %d\n", ans, sum);
        for(int i = sum_cnt - 1; i >= 0; --i)
            printf("%d%c", con[i], i == 0 ? '\n' : ' ');
    }
    return 0;
}

 

2017百度之星1004 度度熊的午饭时光

 

 

 

 

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;

int dp[105][1005];
int score[105], cost[105];
bool re[105][1005];//一切问题的根源在于你的数组开小了

int main()
{
    int CASE;
    scanf("%d", &CASE);
    for(int cas = 1; cas <= CASE; ++cas)
    {
        int pre, n;
        scanf("%d%d", &pre, &n);
        for(int i = 1; i <= n; ++i)
            scanf("%d%d", &score[i], &cost[i]);
        memset(dp, 0, sizeof(dp));
        memset(re, 0, sizeof(re));
        for(int i = 1; i <= n; ++i)
        {

            for(int j = pre; j >= cost[i]; --j)
            {
                if(dp[i - 1][j] <= dp[i - 1][j - cost[i]] + score[i])
                {
                    re[i][j] = 1;
                    dp[i][j] = dp[i - 1][j - cost[i]] + score[i];
                }
                else dp[i][j] = dp[i - 1][j];
            }
            for(int j = cost[i] - 1; j >= 0; --j)
                dp[i][j] = dp[i - 1][j];
        }
        int ans = dp[n][pre];

        if(ans == 0)
        {
            printf("Case #%d:\n", cas);
            printf("0 0\n");
            continue;
        }
        int sum = 10005;
        int sum_cnt = 0;
        int con[105];

        for(int i = n; i >= 1; --i)
            for(int j = pre; j >= 0; --j)
            {
                if(dp[i][j] == ans && re[i][j])
                {
                    int tsum = 0;
                    int cnt = 0;
                    int tmp[105];
                    int x, y;
                    for(x = i, y = j; x >= 1 && y > 0; x -= 1)
                    {
                        if(re[x][y])
                        {
                            y -= cost[x];
                            tsum += x;
                            tmp[cnt++] = x;
                        }
                    }
                    if(y > 0) continue;
                    //for(int t = cnt - 1; t >= 0; --t)
                        //printf("%d%c", tmp[t], t == 0 ? '\n' : ' ');
                    if(tsum < sum)
                    {
                        sum = tsum;
                        sum_cnt = cnt;
                        for(int t = 0; t < cnt; ++t)
                            con[t] = tmp[t];
                    }
                    else if(tsum == sum)
                    {
                        int ok = 0;
                        for(int k = cnt - 1, t = sum_cnt - 1; k >= 0 && t >= 0; --t, --k)
                        {
                            if(con[t] != tmp[k])
                            {
                                ok = con[t] < tmp[k] ? 0 : 1;
                                break;
                            }
                        }
                        if(ok == 1)
                        {
                            sum_cnt = cnt;
                            for(int t = 0; t < cnt; ++t)
                                con[t] = tmp[t];
                        }
                    }
                }
            }
        sum = 0;
        for(int i = 0; i < sum_cnt; ++i) sum += cost[con[i]];
        for(int i = 1; i <= n; ++i)
        {
            if(score[i] != 0 && cost[i] == 0)
            {
                con[sum_cnt++] = i;
            }
        }
        sort(con, con + sum_cnt);
        printf("Case #%d:\n", cas);
        printf("%d %d\n", ans, sum);
        for(int i = 0; i < sum_cnt; ++i)
            printf("%d%c", con[i], i == sum_cnt - 1 ? '\n' : ' ');
    }
    return 0;
}

 

 

 

经过十几发的WA终于AC了这道有毒的题目，最后才发现原来一开始re数组开的是105*105，所以一直没有发现，天哪噜！！！