标签：char pac stdout turn ace min names amp 处理
题意: 给出m个村庄及其距离，给出n个邮局，要求怎么建n个邮局使代价最小。
析：一般的状态方程很容易写出，dp[i][j] = min{dp[i-1][k] + w[k+1][j]}，表示前 j 个村庄用 k 个邮局距离最小，w可以先预处理出来O(n^2)，但是这个方程很明显是O(n^3)，但是因为是POJ，应该能暴过去。。= =，正解应该是对DP进行优化，很容易看出来，w是满足四边形不等式的，也可以推出来 s 是单调的，可以进行优化。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 300 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
int dp[35][maxn], w[maxn][maxn], s[35][maxn];
int a[maxn];
int main(){
while(scanf("%d %d", &n, &m) == 2){
for(int i = 1; i <= n; ++i) scanf("%d", a+i);
for(int i = 1; i <= n; ++i){
w[i][i] = 0;
for(int j = i+1; j <= n; ++j)
w[i][j] = w[i][j-1] + a[j] - a[i+j>>1];
}
memset(dp, INF, sizeof dp);
memset(s, 0, sizeof s);
for(int i = 1; i <= n; ++i) dp[1][i] = w[1][i];
for(int i = 2; i <= m; ++i){
s[i][n+1] = n;
for(int j = n; j >= i; --j)
for(int k = s[i-1][j]; k <= s[i][j+1]; ++k)
if(dp[i][j] > dp[i-1][k] + w[k+1][j]){
dp[i][j] = dp[i-1][k] + w[k+1][j];
s[i][j] = k;
}
}
printf("%d\n", dp[m][n]);
}
return 0;
}
POJ 1160 Post Office (四边形不等式优化DP)
标签：char pac stdout turn ace min names amp 处理
原文地址：http://www.cnblogs.com/dwtfukgv/p/7290790.html
