01背包的路径全纪录

来源：互联网发布：数据抽样方法编辑：程序博客网时间：2024/06/08 19:55

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！！！

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