题目连接:http://ace.delos.com/usacoprob2?a=Vj0fdsir7i8&S=game1
题目大意:博弈游戏。有一个序列,两位选手只能从最左端或者最右端取数,游戏结束时,得分高的获胜。求游戏结束时,第一个选手的最高得分,并输出两位选手的得分。
思路:又是博弈,不过这题挺简单的,用动态规划做。
状态方程:ans[i][j]=max(sum[i][j]-ans[i+1][j],sum[i][j]-ans[i][j-1]); ans[i][j]表示区间[i,j]内,第一位选手的最高得分,sum[i][j]表示区间[i,j]内所有数的和。
ans[1][n]就是第一位选手的最高得分。
代码:
/*
ID: czq1992
LANG: C++
TASK: game1
*/
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int a[110];
int ans[110][110];
int sum[110][110];
int main()
{
int n,i,j;
freopen("game1.in","r",stdin);
freopen("game1.out","w",stdout);
cin>>n;
for(i=1;i<=n;i++) cin>>a[i];
//sum[n-1][n-1]=a[n-1];
for(i=1;i<=n;i++)
{
sum[i][i]=a[i];
for(j=i+1;j<=n;j++)
sum[i][j]=sum[i][j-1]+a[j];
}
//for(i=0;i<n;i++)
// for(j=i;j<n;j++) cout<<sum[i][j]<<endl;
for(i=1;i<=n;i++) ans[i][i]=a[i];
for(i=1;i<=n-1;i++)
for(j=1;j<=n-i;j++)
{
ans[j][j+i]=max(sum[j][j+i]-ans[j+1][j+i],sum[j][j+i]-ans[j][i+j-1]);
}
cout<<ans[1][n]<<" "<<sum[1][n]-ans[1][n]<<endl;
return 0;
}
/*
Compiling...
Compile: OK
Executing...
Test 1: TEST OK [0.000 secs, 3324 KB]
Test 2: TEST OK [0.000 secs, 3324 KB]
Test 3: TEST OK [0.000 secs, 3324 KB]
Test 4: TEST OK [0.000 secs, 3324 KB]
Test 5: TEST OK [0.000 secs, 3324 KB]
Test 6: TEST OK [0.000 secs, 3324 KB]
Test 7: TEST OK [0.000 secs, 3324 KB]
Test 8: TEST OK [0.000 secs, 3324 KB]
Test 9: TEST OK [0.000 secs, 3324 KB]
Test 10: TEST OK [0.000 secs, 3324 KB]
Test 11: TEST OK [0.000 secs, 3324 KB]
Test 12: TEST OK [0.000 secs, 3324 KB]
Test 13: TEST OK [0.000 secs, 3324 KB]
Test 14: TEST OK [0.000 secs, 3324 KB]
Test 15: TEST OK [0.000 secs, 3324 KB]
Test 16: TEST OK [0.000 secs, 3324 KB]
All tests OK.
*/