区间DP-由矩阵连乘所想到的

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1.最优三角剖分

#include <iostream>#include <sstream>#include <stdio.h>#include <algorithm>#include <string.h>#include <stack>#include <queue>#include <set>#include <math.h>#include <time.h>/*=======================================最优三角剖分dp[i][j] = min(dp[i][x] + dp[x][j] + w(i, x, j));i <= x <= j边界条件 dp[i][i + 1] == 0========================================*/#define flush(arr, i) memset(arr, i, sizeof(arr))typedef long long int64;using namespace std;const int MAX_ITEM = 110;const int oo = 0x7fffffff;//const int oo = 0x3f3f3f3f;int dp[MAX_ITEM][MAX_ITEM], wei[MAX_ITEM][MAX_ITEM];int GetWeight(int from, int mid, int to){    return wei[from][mid] + wei[from][to] + wei[mid][to];}int DP(int from, int to){    if(to - from == 1)        return 0;    if(dp[from][to])        return dp[from][to];    int ans = oo;    for(int i = from + 1; i < to; i++)        ans = min(ans, DP(from, i) + DP(i, to) + GetWeight(from, i, to));    return dp[from][to] = ans;}int main(){    freopen("0-data.txt", "r", stdin);    int n;    while(scanf("%d", &n) != EOF)    {        for(int i = 0; i < n; i++)            for(int j = 0; j < n; j++)                scanf("%d", &wei[i][j]);        flush(dp, 0);        printf("%d\n", DP(0, n - 1));    }    return 0;}

2.最大m子段和

#include <iostream>#include <sstream>#include <stdio.h>#include <algorithm>#include <string.h>#include <stack>#include <queue>#include <set>#include <math.h>#include <time.h>/*=======================================最大m子段和9 39 8 7 6 5 4 3 2 117dp[k][p]表示前k个元素划分为p个子段最大值的最小值dp[k][p] = min{dp[k][p], max{dp[x][p - 1] + sum[x + 1][k]}}p - 1 <= x <= k - 1========================================*/#define flush(arr, i) memset(arr, i, sizeof(arr))typedef long long int64;using namespace std;const int MAX_ITEM = 110;const int oo = 0x7fffffff;//const int oo = 0x3f3f3f3f;int dp[MAX_ITEM][MAX_ITEM];int sum[MAX_ITEM];int DP(int k, int p){    if(p == 1)        return dp[k][p] = sum[k];    if(dp[k][p])        return dp[k][p];    int tmp = 0;    dp[k][p] = oo;    for(int i = p - 1; i <= k - 1; i++)    {        tmp = max(DP(i, p - 1), sum[k] - sum[i]);        dp[k][p] = min(dp[k][p], tmp);    }    return dp[k][p];}int main(){    int len, p;    while(scanf("%d%d", &len, &p) != EOF)    {        flush(sum, 0);        flush(dp, 0);        int tmp;        for(int i = 1; i <= len; i++)        {            scanf("%d", &tmp);            sum[i] = sum[i - 1] + tmp;        }        printf("%d\n", DP(len, p));    }    return 0;}

3.石子合并问题

#include <iostream>#include <sstream>#include <stdio.h>#include <algorithm>#include <string.h>#include <stack>#include <queue>#include <set>#include <math.h>#include <time.h>/*=======================================石子合并问题自底向上，求步长为k的时候，步长为k - 1问题的解已知dp[i][j]表示i -> j的最优解 得分最多/最少要加上合并的石子总数作为最优值dp[i][j] = max(dp[i][j], dp[i][k] + dp[k + 1][j] + sum[i][j]);========================================*/#define flush(arr, i) memset(arr, i, sizeof(arr))typedef long long int64;using namespace std;const int MAX_ITEM = 110;const int oo = 0x7fffffff;//const int oo = 0x3f3f3f3f;int dp[MAX_ITEM][MAX_ITEM];int sum[MAX_ITEM];int main(){    int len;    while(scanf("%d", &len) != EOF)    {        for(int i = 1; i <= len; i++) scanf("%d", sum + i);        sum[0] = 0;        for(int i = 1; i <= len; i++)            sum[i] += sum[i - 1];        flush(dp, 0);        for(int sp = 1; sp <= len; sp++)        {            for(int i = 1; i + sp <= len; i++)            {                int j = i + sp;                for(int k = i; k < j; k++)                    dp[i][j] = max(dp[i][j], dp[i][k] + dp[k + 1][j] + sum[j] - sum[i - 1]);                printf("dp[%d][%d] = %d\n", i, j, dp[i][j]);            }        }        printf("%d\n", dp[1][len]);    }    return 0;}

4.最大的算式[TYVJ]

#include <iostream>#include <sstream>#include <stdio.h>#include <algorithm>#include <string.h>#include <stack>#include <queue>#include <set>#include <math.h>#include <time.h>/*=======================================最大的算式dp[k][p] = max(dp[p - 1 ... k - 1][p - 1] * sum[k][p])表示前i部分插入p个乘号获取的最大值========================================*/#define flush(arr, i) memset(arr, i, sizeof(arr))typedef long long int64;using namespace std;const int MAX_ITEM = 110;const int oo = 0x7fffffff;//const int oo = 0x3f3f3f3f;int dp[MAX_ITEM][MAX_ITEM], sum[MAX_ITEM][MAX_ITEM];int num[MAX_ITEM];void getSum(int len){    flush(sum, 0);    flush(dp, 0);    for(int i = 1; i <= len; i++)    {        sum[i][i] = num[i];        for(int j = i + 1; j <= len; j++)            sum[i][j] = sum[i][j - 1] + num[j];    }/*    for(int i = 1; i <= len; i++)        for(int j = i; j <= len; j++)            j == len ? printf("%d\n", sum[i][j]) : printf("%d ", sum[i][j]);*/}int DP(int k, int p){    if(p == 0)    {        dp[k][p] = sum[1][k];        return dp[k][p];    }    if(dp[k][p])        return dp[k][p];    int ans = 0;    for(int i = p - 1; i <= k - 1; i++)        ans = max(ans, DP(i, p - 1) * sum[i + 1][k]);    dp[k][p] = ans;    return ans;}int main(){    int len, p;    while(scanf("%d%d", &len, &p) != EOF)    {        for(int i = 1; i <= len; i++) scanf("%d", num + i);        getSum(len);        printf("%d\n", DP(len, p));    }    return 0;}

5.最大k乘积问题

#include <iostream>#include <sstream>#include <stdio.h>#include <algorithm>#include <string.h>#include <stack>#include <queue>#include <set>#include <math.h>#include <time.h>/*=======================================最大k乘积问题dp[k][p] = max(dp[p - 1 ... k - 1][p - 1] * num[k][p])表示前i项分为j部分的最大乘积========================================*/#define flush(arr, i) memset(arr, i, sizeof(arr))typedef long long int64;using namespace std;const int MAX_ITEM = 110;const int oo = 0x7fffffff;//const int oo = 0x3f3f3f3f;int dp[MAX_ITEM][MAX_ITEM], num[MAX_ITEM][MAX_ITEM];char in[MAX_ITEM];int getNum(char *in){    int len = strlen(in);    flush(num, 0), flush(dp, 0);    for(int i = 1; i <= len; i++)    {        int mul = 0;        for(int j = i; j <= len; j++)        {            mul = mul * 10 + in[j - 1] - '0';            num[i][j] = mul;        }    }/*    for(int i = 1; i <= len; i++)        for(int j = i; j <= len; j++)            j != len ? printf("%d ", num[i][j]) : printf("%d\n", num[i][j]);*/}//前k个分为p部分int DP(int k, int p){    if(p == 1)    {        dp[k][p] = num[1][k];        return num[1][k];    }    if(dp[k][p])        return dp[k][p];    int ans = 0;    for(int i = p - 1; i <= k - 1; i++)        ans = max(ans, DP(i, p - 1) * num[i + 1][k]);    dp[k][p] = ans;    return ans;}int main(){    int p;    while(scanf("%s", in) != EOF)    {        getNum(in);        scanf("%d", &p);        int len = strlen(in);        printf("%d\n", DP(len, p));    }    return 0;}

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