UVALive - 2031 Dance Dance Revolution 三維dp

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題目大意：有一個胖子在玩跳舞機，剛開始的位置在(0,0)，跳舞機有四個方向鍵，上左下右分別對應1,2,3,4.現在有以下規則
1.如果從0位置移動到任意四個位置，消耗能量2
2.如果從非0位置跳到相鄰的位置，如1跳到2或4，消耗能量3
3.如果從非0位置跳到對面的位置，如2跳到4，消耗能量4
4.如果跳同一個位置，消耗能量1
5.兩隻腳不能在同一個位置

解題思路：這題其實很水，直接暴力就可以解決了，討論所有情況，用dp[i][j][k]表示跳第k個數字，左腳在i這個位置，右腳在j這個位置時所消耗的能量，接著分類討論

1.如果其中一隻腳在0上的情況
2.其中一隻腳踩的數字和當前要跳的數字一樣
3.兩隻腳踩的數字和當前的數字不一樣

三種情況，分別在細分即可，具體看代碼

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;#define maxn 50010#define INF 0x3f3f3f3fint dp[5][5][maxn];int seq[maxn];int strength[2] = {4,3};int n;int solve() {    memset(dp, 0x3f, sizeof(dp));    dp[0][seq[0]][0] = dp[seq[0]][0][0] = 2;    for(int i = 1; i < n; i++) {        for(int j = 0; j < 5; j++) {            if(dp[j][seq[i-1]][i-1] != INF) {                if(j == 0) {                    if(seq[i] != seq[i-1])                        dp[seq[i]][seq[i-1]][i] = dp[j][seq[i-1]][i-1] + 2;                    if(seq[i] == seq[i-1])                        dp[j][seq[i-1]][i] = dp[j][seq[i-1]][i-1] + 1;                    else                        dp[j][seq[i]][i] = dp[j][seq[i-1]][i-1] + strength[(seq[i-1] + seq[i]) % 2];                    }                else if(j == seq[i] || seq[i-1] == seq[i])                     dp[j][seq[i-1]][i] = min(dp[j][seq[i-1]][i],dp[j][seq[i-1]][i-1] + 1);                else {                    dp[seq[i]][seq[i-1]][i] = min(dp[j][seq[i-1]][i-1] + strength[(j + seq[i]) % 2], dp[seq[i]][seq[i-1]][i]);                    dp[j][seq[i]][i] = min(dp[j][seq[i-1]][i-1] + strength[(seq[i-1] + seq[i] ) % 2], dp[j][seq[i]][i]);                }            }            if(dp[seq[i-1]][j][i-1] != INF) {                if(j == 0) {                    if(seq[i] != seq[i-1])                        dp[seq[i]][seq[i-1]][i] = dp[seq[i-1]][j][i-1] + 2;                    if(seq[i] == seq[i-1])                        dp[seq[i-1]][j][i] = dp[seq[i-1]][j][i-1] + 1;                    else                        dp[seq[i]][j][i] = dp[seq[i-1]][j][i-1] + strength[(seq[i-1] + seq[i]) % 2];                    }                if(j == seq[i] || seq[i-1] == seq[i])                     dp[seq[i-1]][j][i] = min(dp[seq[i-1]][j][i],dp[seq[i-1]][j][i-1] + 1);                else {                    dp[seq[i]][seq[i-1]][i] = min(dp[seq[i-1]][j][i-1] + strength[(j + seq[i]) % 2], dp[seq[i]][seq[i-1]][i]);                    dp[seq[i]][j][i] = min(dp[seq[i-1]][j][i-1] + strength[(seq[i-1] + seq[i] ) % 2], dp[seq[i]][j][i]);                }            }        }    }    int ans = INF;    for(int i = 0; i < 5; i++)        ans = min(min(ans, dp[seq[n-1]][i][n-1]), dp[i][seq[n-1]][n-1]);    return ans;}int main() {    n = 0;    while(scanf("%d", &seq[n]) != EOF && seq[n++]) {        while(scanf("%d", &seq[n]) && seq[n])            n++;            printf("%d\n", solve());        n = 0;    }    return 0;}

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UVALive - 2031 Dance Dance Revolution 三維dp