本文共 2328 字，大约阅读时间需要 7 分钟。
大意: 给定n*m初始字符矩阵, 有k个新矩阵, 第$i$个矩阵是由初始矩阵区间赋值得到的, 求选择一个新矩阵, 使得其余新矩阵到它距离和最小.
字符集比较小, 可以考虑每次区间覆盖对每个字符的贡献. 区间覆盖转化为差分, 然后前缀和优化.
刚开始辅助数组开多了, 卡内存卡了好久. 看别人代码似乎可以再优化掉两个辅助数组.
#include #include #include #include #include #include #include #include #include #include #include #define REP(i,a,n) for(int i=a;i<=n;++i)#define PER(i,a,n) for(int i=n;i>=a;--i)#define hr putchar(10)#define pb push_back#define lc (o<<1)#define rc (lc|1)#define mid ((l+r)>>1)#define ls lc,l,mid#define rs rc,mid+1,r#define x first#define y second#define io std::ios::sync_with_stdio(false)#define endl '\n'#define DB(a) ({REP(__i,1,n) cout< <<' ';hr;})using namespace std;typedef long long ll;typedef pair pii;const int P = 1e9+7, INF = 0x3f3f3f3f;ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}//headconst int M = 3e5+10, N = 1e3+10;int n, m, k;char s[N][N];int l[M], l2[M], r[M], r2[M], e[M];int c[26][N][N], d[26][N][N];ll sum[N][N], val[26][N][N];void upd(int c[N][N], int l, int r, int l2, int r2, int v) { c[l][r]+=v,c[l][r2+1]-=v,c[l2+1][r]-=v,c[l2+1][r2+1]+=v;}ll get(ll c[N][N], int id) { return c[l2[id]][r2[id]]-c[l[id]-1][r2[id]]-c[l2[id]][r[id]-1]+c[l[id]-1][r[id]-1];}int main() { scanf("%d%d%d", &n, &m, &k); REP(i,1,n) scanf("%s", s[i]+1); REP(i,1,n) REP(j,1,m) s[i][j]-='a'; REP(i,1,k) { char cc; scanf("%d%d%d%d %c", l+i,r+i,l2+i,r2+i,&cc); e[i] = cc-'a'; REP(t,0,25) upd(d[t],l[i],r[i],l2[i],r2[i],-1); REP(t,0,25) upd(c[t],l[i],r[i],l2[i],r2[i],abs(e[i]-t)); } REP(i,1,n) REP(j,1,m) { REP(t,0,25) { d[t][i][j] += d[t][i-1][j]+d[t][i][j-1]-d[t][i-1][j-1]; c[t][i][j] += c[t][i-1][j]+c[t][i][j-1]-c[t][i-1][j-1]; val[t][i][j] = (ll)(k+d[t][i][j])*abs(t-s[i][j])+c[t][i][j]; } sum[i][j] = val[s[i][j]][i][j]+sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]; REP(t,0,25) val[t][i][j]+=val[t][i-1][j]+val[t][i][j-1]-val[t][i-1][j-1]; } ll ans = 1e18; REP(i,1,k) ans=min(ans,sum[n][m]-get(sum,i)+get(val[e[i]],i)); printf("%lld\n", ans);}
转载于:https://www.cnblogs.com/uid001/p/10824711.html