【UVA】11400-Lighting System Design（动态规划）

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这道题感觉状态式不是很好推。。。 WA了好几次是因为排序的时候出问题了。

这道题出在线性结构里了，先说一下最长上升子序列吧。

dp[i]代表了以array[i]结尾的时候，最长子序列长度。

推导的时候，以起点递增的顺序进行推导。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<map>#include<stack>#include<queue>#include<set>#include<ctime>#include<cmath>#include<string>#include<iomanip>#include<climits>#include<cctype>#include<deque>#include<list>#include<sstream>#include<vector>#include<cstdlib>using namespace std;#define _PI acos(-1.0)#define INF (1 << 20)typedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> pill;/*======================================最长上升子序列问题dp[i]代表了以i结尾的序列最长子序列长度======================================*/#define MAXD 10000 + 10int dp[MAXD];int main(){    int array[MAXD];    int n;    while(scanf("%d",&n) != EOF){        memset(dp,0,sizeof(dp));        for(int i = 1 ; i <= n ; i++)            scanf("%d",&array[i]);        dp[1] = 1;        for(int i = 1 ; i < n ; i++)            for(int j = i + 1 ; j <= n ; j ++){                if(array[j] > array[i])                    dp[j] = max(dp[j],dp[i] + 1);                else                    dp[j] = max(dp[j],dp[i]);        }        printf("%d\n",dp[n]);    }    return 0;}

第二个就是最长的公共子序列长度。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<map>#include<stack>#include<queue>#include<set>#include<ctime>#include<cmath>#include<string>#include<iomanip>#include<climits>#include<cctype>#include<deque>#include<list>#include<sstream>#include<vector>#include<cstdlib>using namespace std;#define _PI acos(-1.0)#define INF (1 << 20)typedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> pill;/*======================================最长公公子序列问题======================================*/#define MAXD 1000 + 10int dp[MAXD][MAXD];int main(){    char str1[MAXD];    char str2[MAXD];    while(scanf("%s",str1 + 1) != EOF){        scanf("%s",str2 + 1);        int L1 = strlen(str1 + 1) + 1;        int L2 = strlen(str2 + 1) + 1;        memset(dp,0,sizeof(dp));        for(int i = 1 ; i <= L1 ; i++)            for(int j = 1 ; j <= L2 ; j++){                if(str1[i] == str2[j]){                    dp[i][j] = max(dp[i][j],dp[i - 1][j - 1] + 1);                }                else{                    dp[i][j] = max(dp[i - 1][j],dp[i][j - 1]);                }            }        printf("%d\n",dp[L1 - 1][L2 - 1]);    }    return 0;}

这道题的话，也属于线性推导，d[i]代表了前i种灯泡的最优方略。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<map>#include<stack>#include<queue>#include<set>#include<ctime>#include<cmath>#include<string>#include<iomanip>#include<climits>#include<cctype>#include<deque>#include<list>#include<sstream>#include<vector>#include<cstdlib>using namespace std;#define _PI acos(-1.0)#define INF (1 << 20)typedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> pill;/*======================================*/#define MAXD 100 + 10int n,m;int mat[MAXD][MAXD];int dp[MAXD][MAXD];   /*结点i，j到达最后一行最小的距离*/void init(){    for(int i = 0 ; i < n ;i++)        dp[m - 1][i] = mat[i][m - 1];    return ;}void DP(){    init();    for(int i = m - 2 ; i >= 0; i--)        for(int j = 0 ; j < n ; j++){            int x = dp[i + 1][j % n];            int y = dp[i + 1][(j + 1) % n];            int z = dp[i + 1][(j - 1 + n) % n];        dp[i][j] = min(min(x,y),min(y,z)) + mat[j][i];    }    int ans = INF;    int pos ;    for(int i = 0 ; i < n ; i++){        if(ans > dp[0][i]){            ans = dp[0][i];            pos = i;        }    }    /*根据求出的解打印距离*/    int head = pos;    printf("%d",pos + 1);    for(int i = 1 ; i < m ; i++){         int row[] = {pos - 1,pos + 1,pos};         if(row[0] <  0) row[0] += n;         if(row[1] >= n) row[1] %= n;         sort(row,row + 3);         for(int j = 0 ; j < 3 ; j++){              if(dp[i][row[j]] == dp[i - 1][pos] - mat[pos][i - 1]){                   pos = row[j];                   break;              }         }         printf(" %d",pos + 1);    }    printf("\n%d\n",ans);}int main(){    while(scanf("%d%d",&n,&m) != EOF){        for(int i = 0 ; i < n ; i++)            for(int j = 0 ; j < m ; j++)                scanf("%d",&mat[i][j]);        DP();    }    return 0;}

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