UVA 10891 Game of Sum

来源：互联网发布：靠淘宝运费险赚钱编辑：程序博客网时间：2024/05/17 23:53

题目链接：http://acm.hust.edu.cn/vjudge/problem/19461

题意：一个长度为n的序列，两个人轮流取数。每次玩家可以从左端或者右端取任意数量个数，但不能两端都取。两人都希望自己的得分尽量大。当所有数都取完后，求先手取数之和减去后手取数之和。

思路：dp[l][r]表示取区间[l,r]内的数时，先手之和减去后手之和的最大值。dp[l][r] = max( sum[l,k] - dp[k+1][r] , sum[k,r] - dp[l][k-1] )分别枚举先手从左/右端取走的个数，取最大值，记忆化搜索计算答案。

还有一种方法是参考书上的递推做法,dp[l][r]表示先手取区间[l,r]内的数获得的最大值。dp[l][r] = sum[i,j] - min( dp[l+1][r] , dp[l+2][r] , dp[l+3][r]...dp[r][r] (取左端) , dp[l][l] , dp[l][l+1] , dp[l][l+2] , dp[l][l+3]....dp[l][r-1] (取右端) , 0(全部取完) );

f[l][r] = min( dp[l+1][r] , dp[l+2][r] , dp[l+3][r]...dp[r][r] )

g[l][r] = min( dp[l][l] , dp[l][l+1] , dp[l][l+2] , dp[l][l+3]....dp[l][r-1] )

dp[l][r] = sum - min( 0 , min(f[l][r] , g[l][r]) )

f和g存在递推式

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <cstdlib>#include <iostream>#include <algorithm>#include <stack>#include <map>#include <set>#include <vector>#include <sstream>#include <queue>#include <utility>using namespace std;#define rep(i,j,k) for (int i=j;i<=k;i++)#define Rrep(i,j,k) for (int i=j;i>=k;i--)#define Clean(x,y) memset(x,y,sizeof(x))#define LL long long#define ULL unsigned long long#define inf 0x7fffffff#define mod 100000007const int maxn = 109;int n;LL sum[maxn];int a[maxn];LL dp[maxn][maxn];bool flag[maxn][maxn];void init(){    sum[0] = 0;    rep(i,1,n)    {        scanf("%d",&a[i]);        sum[i] = sum[i-1] + a[i];    }    Clean(flag,false);}LL dfs( int l , int r ){    if ( flag[l][r] ) return dp[l][r];    if ( l == r ) return a[l];    if ( l > r ) return 0;    LL ans = sum[r] - sum[l-1];    rep(i,l,r-1)    {        ans = max( ans , sum[i]-sum[l-1] - dfs(i+1,r) );        ans = max( ans , sum[r] - sum[i] - dfs(l,i) );    }    flag[l][r] = true;    return dp[l][r] = ans;}int main(){    while( cin>>n )    {        if ( !n ) break;        init();        cout<<dfs(1,n)<<endl;    }    return 0;}

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <cstdlib>#include <iostream>#include <algorithm>#include <stack>#include <map>#include <set>#include <vector>#include <sstream>#include <queue>#include <utility>using namespace std;#define rep(i,j,k) for (int i=j;i<=k;i++)#define Rrep(i,j,k) for (int i=j;i>=k;i--)#define Clean(x,y) memset(x,y,sizeof(x))#define LL long long#define ULL unsigned long long#define inf 0x7fffffff#define mod 100000007#define mp make_pair#define fi first#define se second#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define mid int m=(l+r)>>1const int maxn = 109;LL a[maxn];LL sum[maxn];LL dp[maxn][maxn]; //先手取最大值LL f[maxn][maxn];LL g[maxn][maxn];int n;void init(){    sum[0] = 0;    rep(i,1,n)    {        cin>>a[i];        sum[i] = sum[i-1] + a[i];        dp[i][i] = a[i];        f[i][i] = g[i][i] = inf;    }    rep(L,2,n)    {        rep(i,1,n-L+1)        {            int j = i + L - 1;            f[i][j] = min( f[i+1][j] , dp[i+1][j] );            g[i][j] = min( g[i][j-1] , dp[i][j-1] );            dp[i][j] = sum[j] - sum[i-1] - min( 0LL , min( f[i][j] , g[i][j] ) );        }    }    cout<<2*dp[1][n] - sum[n]<<endl;}int main(){    while(cin>>n)    {        if ( !n ) break;        init();    }    return 0;}

0 0