标签：cst i++ style for nes scanf bool printf namespace
还是给你石头n枚，每一枚石头有两个值a和b，每取一个石头，除了这块石头其余所有的石头的a就都减去这个石头的b，问你取了的石头的a的总和最大可以为多少？
先按B从大到小排序
然后DP:
取的话:dp[i][j]=dp[i-1][j-1]+a[i]-b[i]*(j-1) 注意是j-1
不取的话:dp[i][j]=dp[i-1][j];
#include<iostream> #include<cmath> #include<cstdio> #include<sstream> #include<cstdlib> #include<string> #include<string.h> #include<cstring> #include<algorithm> #include<vector> #include<map> #include<set> #include<stack> #include<list> #include<queue> #include<ctime> #include<bitset> #define eps 1e-6 #define INF 0x3f3f3f3f #define PI acos(-1.0) #define ll __int64 #define LL long long #define lson l,m,(rt<<1) #define rson m+1,r,(rt<<1)|1 #define M 1000000007 //#pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; #define Maxn 1100 struct Inf { int a,b; }save[Maxn]; LL dp[Maxn][Maxn]; int n; LL Max(LL a,LL b) { return a>b?a:b; } bool cmp(struct Inf a,struct Inf b) { return a.b>b.b; } int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); while(scanf("%d",&n)&&n) { for(int i=1;i<=n;i++) scanf("%lld%lld",&save[i].a,&save[i].b); sort(save+1,save+n+1,cmp); for(int i=0;i<=n;i++) dp[i][0]=0; for(int i=1;i<=n;i++) { for(int j=1;j<=i;j++) dp[i][j]=max(dp[i-1][j-1]+save[i].a-save[i].b*(j-1),dp[i-1][min(j,i-1)]); //dp[i][j]=Max(dp[i-1][j],dp[i-1][j+1]+save[i].a-save[i].b*j); } LL ans=0; for(int i=1;i<=n;i++) ans=max(ans,dp[n][i]); printf("%lld\n",ans); } return 0; }
Problem 1538 - B - Stones II 贪心+DP
标签：cst i++ style for nes scanf bool printf namespace
原文地址：http://www.cnblogs.com/Aragaki/p/7521575.html
