LightOJ 1031 - Easy Game （区间dp）

题意:

n<=100的整数序列,A和B轮流取数,每个人只能从左端或者右端取任意数量的数
A先手,所有数取完游戏结束,假设两个人都采取最优策略,求SumA−SumB

分析:

区间dp,dp[l][r]:=先手A从s[l...r]序列取得的最大值
转移的话,枚举给后手B留下的序列,最小的那个被减去后就是先手取到的
dp[l][r]:=sum[r]−sum[l−1]−
　　　min{0(取完),min{dp[l+1][r],dp[l+2][r],...,dp[r][r](右边取)},min{dp[l][r−1],dp[l][r−2],...,dp[l][l]}(左边取)};

代码:

////  Created by TaoSama on 2015-11-13//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n, a[105], sum[105], dp[105][105];int dfs(int l, int r) {    int &ret = dp[l][r];    if(ret != INF) return ret;    int left = 0;    for(int i = l; i < r; ++i) left = min(left, dfs(l, i));    for(int i = l + 1; i <= r; ++i) left = min(left, dfs(i, r));    ret = sum[r] - sum[l - 1] - left;    return ret;}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    int kase = 0;    while(t--) {        scanf("%d", &n);        for(int i = 1; i <= n; ++i) scanf("%d", a + i), sum[i] = sum[i - 1] + a[i];        memset(dp, 0x3f, sizeof dp);        printf("Case %d: %d\n", ++kase, 2 * dfs(1, n) - sum[n]);    }    return 0;}

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观察发现可以优化到O(n2)
用右边取f[i][j]:=min{dp[i][j],dp[i+1][j],dp[i+2][j],...,dp[j][j]};
用左边取g[i][j]:=min{dp[i][j],dp[i][j−1],dp[i][j−2],...,dp[i][i]};
转移就很显然了,f[i][j]=min(f[i+1][j],dp[i][j]),g[i][j]=min(g[i][j−1],dp[i][j])

代码:

////  Created by TaoSama on 2015-11-13//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n, a[105], sum[105];int f[105][105], g[105][105], dp[105][105];//f[i][j]:= min{dp[i][j],dp[i+1][j],dp[i+2][j],...,dp[j][j]};//g[i][j]:= min{dp[i][j],dp[i][j-1],dp[i][j-2],...,dp[i][i]};int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    int kase = 0;    while(t--) {        scanf("%d", &n);        for(int i = 1; i <= n; ++i) {            scanf("%d", a + i);            sum[i] = sum[i - 1] + a[i];            dp[i][i] = f[i][i] = g[i][i] = a[i];        }        for(int l = 1; l < n; ++l) {            for(int i = 1; i + l <= n; ++i) {                int j = i + l;                dp[i][j] = sum[j] - sum[i - 1];                dp[i][j] -= min(0, min(f[i + 1][j], g[i][j - 1]));                f[i][j] = min(f[i + 1][j], dp[i][j]);                g[i][j] = min(g[i][j - 1], dp[i][j]);            }        }        printf("Case %d: %d\n", ++kase, 2 * dp[1][n] - sum[n]);    }    return 0;}