dp hdu5433 Xiao Ming climbing

传送门：点击打开链接

题意：告诉你图，然后告诉你初始斗气，你当前的位置，终点位置，每走一步所需要的体力等于两个位置的权值之差除以当前斗气k，然后斗气减一

思路：因为数据很小，所以最简单的方法当然是动态规划去做

设d[i][j][k]表示当前在(i,j)且斗气为k时消耗的最小体力，然后向4个方向转移就行了

#include<map>
#include<set>
#include<cmath>
#include<stack>
#include<queue>
#include<cstdio>
#include<cctype>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define fuck printf("fuck")
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w+",stdout)
using namespace std;
typedef long long LL;

const int MX = 50 + 5;
const int INF = 0x3f3f3f3f;

int dist[][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}};
int Sx, Sy, Ex, Ey;
double d[MX][MX][MX];
char S[MX][MX];

int main() {
int T, m, n, w; //FIN;
scanf("%d", &T);
while(T--) {
scanf("%d%d%d", &m, &n, &w);
for(int i = 1; i <= m; i++) {
scanf("%s", S[i] + 1);
}
scanf("%d%d%d%d", &Sx, &Sy, &Ex, &Ey);
for(int i = 1; i <= m; i++) {
for(int j = 1; j <= n; j++) {
for(int k = 0; k <= w; k++) {
d[i][j][k] = INF;
}
}
}

d[Sx][Sy][w] = 0;
for(int k = w; k >= 2; k--) {
for(int i = 1; i <= m; i++) {
for(int j = 1; j <= n; j++) {
if(d[i][j][k] == INF) continue;
for(int t = 0; t < 4; t++) {
int nx = i + dist[t][0];
int ny = j + dist[t][1];
if(nx < 1 || nx > m || ny < 1 || ny > n || S[nx][ny] == '#') continue;
double value = abs(S[i][j] - S[nx][ny]) * 1.0 / k;
d[nx][ny][k - 1] = min(d[nx][ny][k - 1], d[i][j][k] + value);
}
}
}
}

double ans = INF;
for(int k = 1; k <= w; k++) {
ans = min(ans, d[Ex][Ey][k]);
}
if(ans == INF) printf("No Answer\n");
else printf("%.2lf\n", ans);
}
return 0;
}