Codeforces #803E: Roma and Pokers 题解

这题显然的dp

设dp[i][j]表示当前考虑到第i个位置，胜场比负场多j个能否做到，同时记录转移的坐标方便打印

因为j可能是负的，所以要加一个base

如果当前位是W：dp[i][j]=dp[i-1][j-1]

如果当前位是L：dp[i][j]=dp[i][j+1]

如果当前位是D：dp[i][j]=dp[i-1][j]

如果当前位是？：上面三种情况都考虑一遍

注意：转移的时候abs(j)是不能为k的

#include <cstdio>#include <iostream>#include <cstring>#include <string>#include <cmath>#include <algorithm>#include <cstdlib>#include <utility>#include <map>#include <stack>#include <set>#include <vector>#include <queue>#include <deque>#define x first#define y second#define mp make_pair#define pb push_back#define LL long long#define Pair pair<bool,pair<int,int> >#define LOWBIT(x) x & (-x)using namespace std;const int zero_stand=1500;const int MOD=1e9+7;const int INF=0x7ffffff;const int magic=348;Pair dp[1048][3048];int n,k;char s[1048];void print(int x,int y){if (dp[x][y].y.y!=-1) print(x-1,dp[x][y].y.y);if (dp[x][y].y.x==1) printf("W");if (dp[x][y].y.x==2) printf("L");if (dp[x][y].y.x==3) printf("D");}int main (){int i,j;scanf("%d%d%s",&n,&k,s+1);for (i=1;i<=n;i++)for (j=zero_stand-k-10;j<=zero_stand+k+10;j++)dp[i][j]=mp(false,mp(0,0));if (s[1]=='W') dp[1][zero_stand+1]=mp(true,mp(1,-1));if (s[1]=='L') dp[1][zero_stand-1]=mp(true,mp(2,-1));if (s[1]=='D') dp[1][zero_stand]=mp(true,mp(3,-1));if (s[1]=='?'){dp[1][zero_stand+1]=mp(true,mp(1,-1));dp[1][zero_stand-1]=mp(true,mp(2,-1));dp[1][zero_stand]=mp(true,mp(3,-1));}for (i=2;i<=n;i++)for (j=zero_stand-k;j<=zero_stand+k;j++){if ((j==zero_stand-k || j==zero_stand+k) && i!=n) continue;if (s[i]=='W')if (dp[i-1][j-1].x && j-1!=zero_stand-k){dp[i][j]=mp(true,mp(1,j-1));continue;}if (s[i]=='L')if (dp[i-1][j+1].x && j+1!=zero_stand+k){dp[i][j]=mp(true,mp(2,j+1));continue;}if (s[i]=='D')if (dp[i-1][j].x){dp[i][j]=mp(true,mp(3,j));continue;}if (s[i]=='?'){if (dp[i-1][j-1].x && j-1!=zero_stand-k){dp[i][j]=mp(true,mp(1,j-1));continue;}if (dp[i-1][j+1].x && j+1!=zero_stand+k){dp[i][j]=mp(true,mp(2,j+1));continue;}if (dp[i-1][j].x){dp[i][j]=mp(true,mp(3,j));continue;}}}if (!dp[n][zero_stand-k].x && !dp[n][zero_stand+k].x){printf("NO\n");return 0;}if (dp[n][zero_stand-k].x) print(n,zero_stand-k); else print(n,zero_stand+k);printf("\n");return 0;}