UVA 5875 DP

题意：给你一堆二维点，每个点有一些分数。

现在要从点(0 , 0 )出发，只能从标号小的点走到大的点，每个人有一个走的距离的限制，问最后能拿到的最高的分数，当然这个人从（0 ， 0）出发还得回到( 0 , 0 )。

思路：用dp[i][j]表示在点i分数为j的最短距离，转移方程就是dp[i][j] = min(dp[i][j] , dp[k][j - w[k]] + Map[i][k]) 。

CODE：

#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <vector>
#include <iomanip>
#include <cstring>
#include <iostream>
#include <algorithm>
#define Max 2505
#define FI first
#define SE second
#define ll __int64
#define PI acos(-1.0)
#define inf 0x3fffffff
#define LL(x) ( x << 1 )
#define bug puts("here")
#define PII pair<int,int>
#define RR(x) ( x << 1 | 1 )
#define mp(a,b) make_pair(a,b)
#define mem(a,b) memset(a,b,sizeof(a))
#define REP(i,s,t) for( int i = ( s ) ; i <= ( t ) ; ++ i )

using namespace std;

#define N 333
ll x
, y
, w
;
int n ;
double Map

;
double dp
[11111] ;
double getD(int i , int j) {
return sqrt(1.0 * (x[i] - x[j]) * (x[i] - x[j]) + 1.0 * (y[i] - y[j]) * (y[i] - y[j])) ;
}
int Asum = 0 ;
int main() {
int ca = 0 ;
while(cin >> n , n ) {
Asum = 0 ;
for (int i = 1 ; i <= n ; i ++ ) {
scanf("%I64d%I64d%I64d",&x[i] ,&y[i],&w[i]) ;
Asum += w[i] ;
}
n ++ ;
x[0] = y[0] = x
= y
= w[0] = w
= 0 ;
for (int i = 0 ; i <= n ; i ++ ) {
for (int j = 0 ; j <= n ; j ++ ) {
Map[i][j] = getD(i , j) ;
}
}
string name ;
int lim ;
printf("Race %d\n",++ca) ;
while(cin >> name >> lim) {
if(name == "#")break ;
for (int i = 0 ; i <= n ; i ++ ) {
for (int j = 0 ; j <= Asum ; j ++ )dp[i][j] = inf ;
dp[0][0] = 0 ;
}
for (int i = 1 ; i <= n ; i ++ ) {
for (int j = 0 ; j <= Asum ; j ++ ) {
for (int k = 0 ; k < i ; k ++ ) {
if(w[k] > j)continue ;
dp[i][j] = min(dp[i][j] , dp[k][j - w[k]] + Map[k][i]) ;
}
}
}
int ans = 0 ;
for (int i = 0 ; i <= Asum ; i ++ ) {
if(dp
[i] <= lim)ans = max(ans , i) ;
}
printf("%s: %d\n",name.c_str() ,ans) ;
}
}
return 0 ;
}

/*

5
750 -800 30
1500 0 50
750 750 60
-1250 750 70
-1000 -500 50
Chris 7000
Karl 6500
Tania 5000
# 0
4
500 0 10
0 500 10
-500 0 10
0 -500 10
Hanny 2100
Lizzie 1800
# 0
0

*/