uva1347 tour

#include<bits/stdc++.h>
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define MS0(a) memset(a,0,sizeof(a))

using namespace std;

typedef long long ll;
const int maxn=1100;
const int INF=1<<29;
const double EPS=0.0000001;

double dp[maxn][maxn];
int n;
struct Point
{
    int x,y;
};Point p[maxn];

double dist(Point A,Point B)
{
    double tx=A.x-B.x,ty=A.y-B.y;
    return sqrt(tx*tx+ty*ty);
}

int main()
{
    freopen("in.txt","r",stdin);
    while(cin>>n){
        REP(i,1,n) scanf("%d%d",&p[i].x,&p[i].y);
        REP(i,1,n) REP(j,1,n) dp[i][j]=INF*1.0;
        dp[1][1]=0;
        REP(i,1,n){
            REP(j,1,n){
                if(i>=j){
                    if(i+1!=j||i+1==n) dp[i+1][j]=min(dp[i+1][j],dp[i][j]+dist(p[i],p[i+1]));
                    if(i!=n) dp[i][i+1]=min(dp[i][i+1],dp[i][j]+dist(p[j],p[i+1]));
                    else dp[i][i]=min(dp[i][i],dp[i][j]+dist(p[j],p[i]));
                }
                else{
                    if(j+1!=i||j+1==n) dp[i][j+1]=min(dp[i][j+1],dp[i][j]+dist(p[j],p[j+1]));
                    if(j!=n) dp[j+1][j]=min(dp[j+1][j],dp[i][j]+dist(p[i],p[j+1]));
                    else dp[j][j]=min(dp[j][j],dp[i][j]+dist(p[i],p[j]));
                }
            }
        }
        /*
        REP(i,1,n){
            REP(j,1,n){
                printf("%15.2f ",dp[i][j]);
            }
            puts("");
        }
        */
        printf("%.2f
",dp[n][n]);
    }
    return 0;
}
/**
dp决策的考虑的两种方式:
用一般方法考虑状态转移:从哪里转移过来。
用刷表法考虑状态转移:转移到哪里去。
*/
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没有AC不了的题,只有不努力的ACMER!
原文地址:https://www.cnblogs.com/--560/p/5063336.html