哎，注意細節啊，，，，，，，思維的嚴密性。。。。。

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11699193

2014-09-22 08:46:42

Accepted

5037

796MS

1864K

2204 B

G++

czy

Frog

Time Limit: 3000/1500 MS (Java/Others)????Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 454????Accepted Submission(s): 96

Problem Description Once upon a time, there is a little frog called Matt. One day, he came to a river.
?? The river could be considered as an axis.Matt is standing on the left bank now (at position 0). He wants to cross the river, reach the right bank (at position M). But Matt could only jump for at most L units, for example from 0 to L.
As the God of Nature, you must save this poor frog.There are N rocks lying in the river initially. The size of the rock is negligible. So it can be indicated by a point in the axis. Matt can jump to or from a rock as well as the bank.
?? You don't want to make the things that easy. So you will put some new rocks into the river such that Matt could jump over the river in maximal steps.And you don't care the number of rocks you add since you are the God.
?? Note that Matt is so clever that he always choose the optimal way after you put down all the rocks. Input The first line contains only one integer T, which indicates the number of test cases.
?? For each test case, the first line contains N, M, L (0<=N<=2*10^5,1<=M<=10^9, 1<=L<=10^9).
?? And in the following N lines, each line contains one integer within (0, M) indicating the position of rock. Output For each test case, just output one line “Case #x: y", where x is the case number (starting from 1) and y is the maximal number of steps Matt should jump. Sample Input 2 1 10 5 5 2 10 3 3 6 Sample Output Case #1: 2 Case #2: 4 Source 2014 ACM/ICPC Asia Regional Beijing Online Recommend hujie???|???We have carefully selected several similar problems for you:??5041?5040?5039?5038?5036?

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1 #include<iostream> 2 #include<cstring> 3 #include<cstdlib> 4 #include<cstdio> 5 #include<algorithm> 6 #include<cmath> 7 #include<queue> 8 #include<map> 9 #include<string> 10 11 #define N 200005 12 #define M 15 13 #define mod 10000007 14 //#define p 10000007 15 #define mod2 100000000 16 #define ll long long 17 #define LL long long 18 #define maxi(a,b) (a)>(b)? (a) : (b) 19 #define mini(a,b) (a)<(b)? (a) : (b) 20 21 using namespace std; 22 23 int T; 24 int n; 25 int m,l; 26 int dp[N]; 27 int p[N]; 28 29 void ini() 30 { 31 memset(dp,0,sizeof(dp)); 32 scanf("%d%d%d",&n,&m,&l); 33 for(int i=1;i<=n;i++){ 34 scanf("%d",&p[i]); 35 } 36 sort(p+1,p+1+n); 37 p[n+1]=m; 38 } 39 40 41 void solve() 42 { 43 int sh; 44 int te; 45 int now; 46 int end; 47 int d; 48 int tnow; 49 now=0; 50 d=1; 51 for(int i=1;i<=n+1;){ 52 dp[i]=dp[i-1]; 53 te=(p[i]-now); 54 sh=te/(l+1); 55 dp[i]+=sh*2+1; 56 if(te%(l+1)!=0){ 57 //dp[i]--; 58 // if(sh!=0 && te%(l+1)<d){ 59 // dp[i]--; 60 // tnow=now+(sh-1)*(l+1)+d; 61 // end=tnow+l; 62 // now=p[i]; 63 64 // } 65 // else{ 66 now=now+sh*(l+1); 67 tnow=now; 68 end=tnow+l; 69 now=p[i]; 70 // } 71 72 i++; 73 while(i<=n+1 && p[i]<=end){ 74 dp[i]=dp[i-1]; 75 now=p[i]; 76 i++; 77 } 78 d=l+1-(now-tnow); 79 } 80 else{ 81 dp[i]--; 82 tnow=now+(sh-1)*(l+1)+d; 83 end=tnow+l; 84 now=p[i]; 85 i++; 86 while(i<=n+1 && p[i]<=end){ 87 dp[i]=dp[i-1]; 88 now=p[i]; 89 i++; 90 } 91 d=l+1-(now-tnow); 92 } 93 } 94 } 95 96 void out() 97 { 98 printf("%d\n",dp[n+1]); 99 } 100 101 int main() 102 { 103 //freopen("data.in","r",stdin); 104 //freopen("data.out","w",stdout); 105 scanf("%d",&T); 106 for(int cnt=1;cnt<=T;cnt++) 107 // while(T--) 108 // while(scanf("%d%d",&n,&m)!=EOF) 109 { 110 // if(n==0 && m==0) break; 111 printf("Case #%d: ",cnt); 112 ini(); 113 solve(); 114 out(); 115 } 116 117 return 0; 118 }

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轉載于:https://www.cnblogs.com/njczy2010/p/3985370.html

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