51NOD1337 翻转游戏 【贪心】

传送门

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显然当没有'?'时,模拟一遍即可当位置i是'?'时,如果后续的关卡中,位置i需要关灯,且当前关卡必须关掉一些灯，则顺手关掉开灯也同理

#include<stdio.h>#include<bits/stdc++.h>#define ll long long#define pii pair<int,int>#define pll pair<ll,ll>#define MEM(a,x) memset(a,x,sizeof(a))#define lowbit(x) ((x)&-(x))using namespace std;const int inf=1e9+7;const int N = 50+5;char str[N][N];int to[N][N];//0:必须关,1必须开,2尽量关,3尽量开int now[N];void initTo(int n,int m){    for(int i=0;i<n;++i){        for(int j=0;j<m;++j){            switch(str[i][j]){            case '-':                to[i][j]=0;                break;            case '+':                to[i][j]=1;                break;            case '?':                to[i][j]=2;                for(int k=i+1;k<n;++k){                    if(str[k][j]!='?'){                        to[i][j]=2+(str[k][j]=='+');                        break;                    }                }                break;            }        }    }}int slove(int n,int m){    initTo(n,m);    int ans=n;    MEM(now,0);    for(int i=0;i<n;++i){        bool mustOpen=0;        bool mustClose=0;        for(int j=0;j<m;++j){            if(to[i][j]==0&&now[j]==1){                mustClose=1;            }            if(to[i][j]==1&&now[j]==0){                mustOpen=1;            }        }        ans+=mustOpen+mustClose;        for(int j=0;j<m;++j){            if(to[i][j]<2){                now[j]=to[i][j];            }            if(mustClose&&to[i][j]==2){                now[j]=0;            }            if(mustOpen&&to[i][j]==3){                now[j]=1;            }        }    }    return ans;}int main(){   // freopen("/home/lu/code/r.txt","r",stdin);    //freopen("/home/lu/code/w.txt","w",stdout);    int T;    scanf("%d",&T);    while(T--){        int n,m;        scanf("%d%d",&n,&m);        for(int i=0;i<n;++i){            scanf("%s",str[i]);        }        printf("%d\n",slove(n,m));    }    return 0;}