題目

輸入格式

輸出格式

僅包含一個數字，表示在執行最優策略時，人物活著走出迷宮的概率。四舍五入保留3位小數。

輸入樣例

4 3 3 2
.$.
A#B
A#C
@@@
143 37 335 85 95 25 223 57

輸出樣例

0.858

提示

題解

毒瘤dp題
我們設\(f[x][y][s][h]\)表示從點\((x,y)\)出發，所有陷阱狀態為\(s\)，生命值為\(h\)，存活的期望概率
我們枚舉鄰點，選擇存活概率最大的作為當前\(f\)的值

除了墻，有以下情況：
①如果是空地或者終點，直接轉移\(f[nx][ny][s][h]\)
②如果是陷阱：
1、如果陷阱已知
無害則同空地的轉移
有害則轉移的同時\(h - 1\)

至於g數組的預處理，對於每種s，枚舉未知位置的子集，將各種情況有害的加到對應陷阱去，然後除以總值

為什麽換一個搜索順序才能A？？？

#include<iostream>
#include<cstdio>
#include<cctype>
#include<cstring>
 

#include<algorithm>
#define REP(i,n) for (int i = 1; i <= (n); i++)
using namespace std;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == ‘-‘) flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3
 
) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
double f[32][32][250][6],g[250][6],p[100];
int vis[32][32][250][6],bin[10];
int n,m,K,H,Sx,Sy,X[4] = {1,0,-1,0},Y[4] = {0,-1,0,1};
int G[32][32];
void init(){
    n = read(); m = read(); K = read(); H = read();
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++){
            char c = getchar();
            while (!isprint(c)) c = getchar();
            if (c == ‘.‘) G[i][j] = 0;
            else if (c == ‘#‘) G[i][j] = -1;
            else if (c == ‘$‘) G[i][j] = 0,Sx = i,Sy = j;
            else if (c == ‘@‘) G[i][j] = -2;
            else G[i][j] = c - ‘A‘ + 1;
        }
    //REP(i,n) {REP(j,m) printf("%d ",G[i][j]); puts("");}
    bin[0] = 1;
    for (int i = 1; i <= K; i++) bin[i] = bin[i - 1] * 3;
    //REP(i,K) printf("%d ",bin[i]); puts("");
    int maxv = (1 << K) - 1,maxp = bin[K] - 1;
    for (int s = 0; s <= maxv; s++) p[s] = read();
    for (int s = 0; s <= maxp; s++){
        int e = 0,t = 0; double sum = 0;
        for (int i = s,j = 1; j <= K; j++,i /= 3){
            if (i % 3 == 0) t |= (1 << j - 1);
            else if (i % 3 == 2) e |= (1 << j - 1);
        }
        for (int i = t; ; i = (i - 1) & t){
            int to = (e | i);
            sum += p[to];
            for (int j = 1; j <= K; j++)
                if (to & (1 << j - 1)) g[s][j] += p[to];
            if (!i) break;
        }
        for (int i = 1; i <= K; i++)
            g[s][i] /= sum;
    }
}
double F(int x,int y,int s,int h){
    if (vis[x][y][s][h]) return f[x][y][s][h];
    if (h == 0){
        vis[x][y][s][h] = 1;
        return f[x][y][s][h] = 0;
    }
    if (G[x][y] == -2){
        vis[x][y][s][h] = 1;
        return f[x][y][s][h] = 1;
    }
    vis[x][y][s][h] = 1;
    double& ff = f[x][y][s][h];
    ff = 0;
    int nx,ny;
    for (int k = 0; k < 4; k++){
        nx = x + X[k];
        ny = y + Y[k];
        if (nx < 1 || ny < 1 || nx > n || ny > m || G[nx][ny] == -1) continue;
        if (G[nx][ny] == 0 || G[nx][ny] == -2){
            ff = max(ff,F(nx,ny,s,h));
        }
        else {
            int t = G[nx][ny];
            if ((s / bin[t - 1]) % 3 == 1) ff = max(ff,F(nx,ny,s,h));
            else if ((s / bin[t - 1]) % 3 == 2) ff = max(ff,F(nx,ny,s,h - 1));
            else {
                ff = max(ff,g[s][t] * F(nx,ny,s + 2 * bin[t - 1],h - 1) + (1 - g[s][t]) * F(nx,ny,s + bin[t - 1],h));
            }
        }
    }
    return ff;
}
int main(){
    init();
    if (n == 0) return 0;
    else printf("%.3lf\n",F(Sx,Sy,0,H));
    return 0;
}

BZOJ2246 [SDOI2011]迷宮探險 【記憶化搜索dp + 概率】