经典DP 二维换一维

HDU 1024  Max Sum Plus Plus

// dp[i][j] = max(dp[i][j-1], dp[i-1][t]) + num[j]
// pre[j-1] 存放dp[i-1][t] 里的 (1<=t<=j-1)最大值。
//dp[j] = max(dp[j-1], pre[j-1]) + num[j];
 
#include <stdio.h>
#include <string.h>
#include <iostream>
#define inf 100000000
#define maxn 1000010
using namespace std;
 
int pre[maxn];
int num[maxn];
int dp[maxn];
 
int main() {
    //freopen("in.cpp", "r", stdin);
    int m, n;
    while(~scanf("%d%d", &m, &n)) {
        for (int i=1; i<=n; ++i)
            scanf("%d", &num[i]);
 
        pre[0] = 0;
        dp[0] = 0;
        memset(pre, 0, sizeof(pre));
 
        int maxx = -inf;
        for (int i=1; i<=m; ++i) {
            maxx = -inf;
            for (int j=i; j<=n; ++j) {
                dp[j] = max(dp[j-1], pre[j-1]) + num[j];
                //pre[j-1] = max(dp[j-1], pre[j-2]);
                pre[j-1] = maxx;
                maxx = max(maxx, dp[j]);
            }
        }
        printf("%d\n", maxx);
    }
    return 0;
}

POJ 1322 Chocolate

二维：

#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;
 
double dp[10010][210];
 
bool judge(int n, int m, int k) {
    if (m > n || m > k) return false;
    if ((m+n)%2) return false;
    return true;
}
 
int main() {
    int k, n, m;
   // freopen("in.cpp", "r", stdin);
    while (~scanf("%d", &k) && k) {
        scanf("%d%d", &n, &m);
 
        if (judge(n, m, k) == false) {
            printf("0.000\n");
            continue;
        }
 
        if (n>10000) {
            if (n%2) n = 10003;
            else n = 10004;
        }
        memset(dp, 0, sizeof(dp));
        dp[0][0] = 1;
 
       for (int i=1; i<=n; ++i) {
            for (int j=0; j<=i && j<=k; ++j) {
                if ((i+j)%2) continue;
                if(j>0) dp[i][j] += dp[i-1][j-1]*(1-(j-1)*1.0/k);
                if(j<k) dp[i][j] += dp[i-1][j+1]*(1.0*(j+1)/k);
            }
        }
 
        printf("%.3f\n", dp[n][m]);
    }
    return 0;
}

一维：

#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;
 
double dp[1000010];
double pre[1000010];
 
bool judge(int n, int m, int k) {
    if (m > n || m > k) return false;
    if ((m+n)%2) return false;
    return true;
}
 
int main() {
    int k, n, m;
    //freopen("in.cpp", "r", stdin);
    while (~scanf("%d", &k) && k) {
        scanf("%d%d", &n, &m);
 
        if (judge(n, m, k) == false) {
            printf("0.000\n");
            continue;
        }
 
        if (n>10000) {
            if (n%2) n = 10003;
            else n = 10004;
        }
        memset(dp, 0, sizeof(dp));
        memset(pre, 0, sizeof(pre));
        pre[0] = 1;
 
       for (int i=1; i<=n; ++i) {
            //memset(dp, 0, sizeof(dp));
            for (int j=0; j<=i && j<=k; ++j) {
                dp[j] = 0;
                if ((i+j)%2) continue;
                if(j>0) dp[j] += pre[j-1]*(1-(j-1)*1.0/k);
                if(j<k) dp[j] += pre[j+1]*(1.0*(j+1)/k);
            }
            for (int j=0; j<=i && j<=k; ++j) {
                pre[j] = dp[j];
            }
        }
 
       printf("%.3f\n", pre[m]);
    }
    return 0;
}