hdu 4412 Sky Soldiers DP

动态规划，主要是用单调性求区间的最小期望。

代码如下：

 

 1 #include<iostream>

 2 #include<stdio.h>

 3 #include<algorithm>

 4 #include<iomanip>

 5 #include<cmath>

 6 #include<cstring>

 7 #include<vector>

 8 #include<map>

 9 #define MAX 1004

10 #define inf 1<<22

11 using namespace std;

12 double dp[MAX][MAX],b,add,l,r;

13 map<int,double>q;

14 struct node

15 {

16     int x;

17     double p;

18 }an[MAX];

19 void init(int m)

20 {

21     int i,j;

22     for(i=0;i<=1000;i++)

23         for(j=0;j<=50;j++)

24             dp[i][j]=inf;

25     for(i=0;i<=m;i++)

26         for(j=i;j<=m;j++)

27             dp[i][j]=0.0;

28 }

29 int main(){

30     int n,m,k,i,j,a;

31     while(scanf("%d%d",&n,&m)&&(n+m)){

32         q.clear();

33         for(i=0;i<n;i++){

34             scanf("%d",&k);

35             for(j=0;j<k;j++){

36                 scanf("%d%lf",&a,&b);

37                 q[a]+=b;

38             }

39         }

40         n=0;

41         for(map<int,double>::iterator it=q.begin();it!=q.end();it++){

42             an[++n].x=it->first;

43             an[n].p=it->second;

44         }

45         init(m);

46         for(i=1;i<=n;i++){

47             for(j=1;j<=m&&j<=i;j++){

48                 dp[i][j]=inf;

49                 double pre=0.0,suf=0.0;

50                 int cur=i;add=0.0;

51                 for(k=i-1;k>=j-1;k--){

52                     if(k+1!=i){

53                         pre+=an[k+1].p;

54                         add+=an[k+1].p*(an[cur].x-an[k+1].x);

55                     }

56                     while(an[cur].p+suf<pre){

57                         suf+=an[cur].p;

58                         cur--;

59                         pre-=an[cur].p;

60                         int t=an[cur+1].x-an[cur].x;

61                         add-=an[cur].p*t;

62                         add+=an[cur+1].p*t;

63                         add-=pre*t;

64                         add+=(suf-an[cur+1].p)*t;

65                     }

66                     dp[i][j]=min(dp[i][j],dp[k][j-1]+add);

67                 }

68             }

69         }

70         printf("%.2lf\n",dp[n][m]);

71     }

72     return 0;

73 }

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