【HDU5074】Hatsune Miku(简单DP)

记录一个菜逼的成长。。

题解摘自：http://blog.csdn.net/sr_19930829/article/details/40392311?utm_source=tuicool&utm_medium=referral
分情况讨论：

当note[i] >0 （i>=2)时：

当note[i-1]>0时： dp[i][note[i]]=dp[i-1][note[i-1]]+score[note[i-1]][note[i]];

当note[i-1]<0时： dp[i][note[i]]=max(dp[i][note[i]],dp[i-1][j]+score[j][note[i]]); 1<=j<=m， 枚举第i-1个音符是第几种音符

当note[i]<0 (i>=2)时：

当note[i-1]>0时：dp[i][j]=max(dp[i][j],dp[i-1][note[i-1]]+score[note[i-1]][j]); 1<=j<=m, 枚举第i个音符是第几种音符

当note[i-1]<0时：dp[i][j]=max(dp[i][j],dp[i-1][k]+score[k][j]); 1<=j,k<=m, 枚举第i个音符和第i-1个音符是第几种音符

PS:虽然一看我也知道是简单DP，但状态定义不是很准确/(ㄒoㄒ)/~~，还得多学习。

#include <cstdio>#include <iostream>#include <cstring>#include <string>#include <algorithm>#include <cstdlib>#include <vector>#include <set>#include <map>#include <queue>#include <stack>#include <list>#include <deque>#include <cctype>#include <bitset>#include <cmath>using namespace std;#define ALL(v) (v).begin(),(v).end()#define cl(a) memset(a,0,sizeof(a))#define bp __builtin_popcount#define pb push_back#define fin freopen("D://in.txt","r",stdin)#define fout freopen("D://out.txt","w",stdout)#define lson t<<1,l,mid#define rson t<<1|1,mid+1,r#define seglen (node[t].r-node[t].l+1)typedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> PII;typedef pair<LL,LL> PLL;typedef vector<PII> VPII;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;template <typename T>inline void read(T &x){    T ans=0;    char last=' ',ch=getchar();    while(ch<'0' || ch>'9')last=ch,ch=getchar();    while(ch>='0' && ch<='9')ans=ans*10+ch-'0',ch=getchar();    if(last=='-')ans=-ans;    x = ans;}inline bool DBread(double &num){    char in;double Dec=0.1;    bool IsN=false,IsD=false;    in=getchar();    if(in==EOF) return false;    while(in!='-'&&in!='.'&&(in<'0'||in>'9'))        in=getchar();    if(in=='-'){IsN=true;num=0;}    else if(in=='.'){IsD=true;num=0;}    else num=in-'0';    if(!IsD){        while(in=getchar(),in>='0'&&in<='9'){            num*=10;num+=in-'0';}    }    if(in!='.'){        if(IsN) num=-num;            return true;    }else{        while(in=getchar(),in>='0'&&in<='9'){                num+=Dec*(in-'0');Dec*=0.1;        }    }    if(IsN) num=-num;    return true;}template <typename T>inline void write(T a) {    if(a < 0) { putchar('-'); a = -a; }    if(a >= 10) write(a / 10);    putchar(a % 10 + '0');}/******************head***********************/const int maxn = 100 + 10;int a[maxn][maxn],dp[maxn][maxn],note[maxn];int main(){    int T;read(T);    while(T--){        cl(dp);        int n,m;        read(n),read(m);        for( int i = 1; i <= m; i++ ){            for( int j = 1; j <= m; j++ ){                read(a[i][j]);            }        }        for( int i = 1; i <= n; i++ ){            read(note[i]);        }        for( int i = 2; i <= n; i++ ){            if(note[i] > 0){                if(note[i-1] > 0){                    dp[i][note[i]] = dp[i-1][note[i-1]] + a[note[i-1]][note[i]];                }                else {                    for( int j = 1; j <= m; j++ ){                        dp[i][note[i]] = max(dp[i][note[i]],dp[i-1][j] + a[j][note[i]]);                    }                }            }            else {                if(note[i-1] > 0){                    for( int j = 1; j <= m; j++ ){                        dp[i][j] = max(dp[i][j] , dp[i-1][note[i-1]] + a[note[i-1]][j]);                    }                }                else {                    for( int j = 1; j <= m; j++ ){                        for( int k = 1; k <= m; k++ ){                            dp[i][k] = max(dp[i][k],dp[i-1][j] + a[j][k]);                        }                    }                }            }        }        int ans = 0;        for( int i = 1; i <= m; i++ ){            ans = max(ans,dp[n][i]);        }        printf("%d\n",ans);    }    return 0;}