P8675 [蓝桥杯 2018 国 B] 搭积木

合集 - 动态规划dp(55)

1.P8784 [蓝桥杯 2022 省 B] 积木画2024-02-172.P9847 [ICPC2021 Nanjing R] Crystalfly2024-02-183.G. Vlad and Trouble at MIT2024-02-204.P2899 [USACO08JAN] Cell Phone Network G2024-02-205.P5322 [BJOI2019] 排兵布阵2024-03-036.P2946 [USACO09MAR] Cow Frisbee Team S2024-03-037.P1156 垃圾陷阱2024-03-038.P1064 [NOIP2006 提高组] 金明的预算方案2024-03-049.P2466 [SDOI2008] Sue 的小球2024-03-0410.P5020 [NOIP2018 提高组] 货币系统2024-03-0411.P1757 通天之分组背包2024-03-0412.P2135 方块消除2024-03-0613.D - String Bags2024-03-1014.P4395 [BOI2003] Gem 气垫车2024-03-1015.A - A Multiply2024-03-2416.P2642 双子序列最大和2024-03-2517.P1121 环状最大两段子段和2024-03-2518.P2422 良好的感觉2024-03-2519.P2854 [USACO06DEC] Cow Roller Coaster S2024-03-2620.C. Theofanis' Nightmare2024-03-2721.D. Birthday Gift2024-03-2722.P1284 三角形牧场2024-03-2823.P1353 [USACO08JAN] Running S2024-03-2824.P8736 [蓝桥杯 2020 国 B] 游园安排2024-03-3025.P1470 [USACO2.3] 最长前缀 Longest Prefix2024-04-0126.P7859 [COCI2015-2016#2] GEPPETTO2024-04-0527.P1435 [IOI2000] 回文字串2024-05-0828.P2340 [USACO03FALL] Cow Exhibition G2024-05-1329.P1854 花店橱窗布置2024-05-1330.P3147 [USACO16OPEN] 262144 P2024-05-1331.Game on Tree2024-05-1432.P2938 [USACO09FEB] Stock Market G2024-05-1833.P9691 [GDCPC2023] Base Station Construction2024-05-1834.P8624 [蓝桥杯 2015 省 AB] 垒骰子2024-05-2035.E. Money Buys Happiness2024-05-22

36.P8675 [蓝桥杯 2018 国 B] 搭积木2024-05-22

37.P8774 [蓝桥杯 2022 省 A] 爬树的甲壳虫2024-05-2238.P8806 [蓝桥杯 2022 国 B] 搬砖2024-05-2339.P5662 [CSP-J2019] 纪念品2024-05-2440.B. Mashmokh and ACM2024-06-0841.C. Minimizing the Sum2024-06-1242.H. Don't Blame Me2024-06-1343.D. Bicolorings2024-06-1444.D. Armchairs2024-06-1845.C. Chef Monocarp2024-06-1846.D. Invertible Bracket Sequences2024-06-2047.P1351 [NOIP2014 提高组] 联合权值2024-06-2648.D. World is Mine2024-07-0249.F. Feed Cats2024-07-0750.P3522 [POI2011] TEM-Temperature2024-07-2151.P3572 [POI2014] PTA-Little Bird2024-07-2252.P3089 [USACO13NOV] Pogo-Cow S2024-07-2253.E - Maximum Glutton2024-07-2854.Projects2024-07-2855.Elevator Rides2024-07-28

收起

原题链接

题解

1.请务必读清题干意思
2.如果以最顶端积木的位置为状态，是可以穷尽所有情况的，则状态为 $dp[i][l][r]$ ，最顶端第 $i$ 层只在区间 $[l,r]$ 内连续放置积木有几种方法
3.状态转移方程 $dp[i][l][r]=\sum _1^l\sum _r^{m}dp[i+1][x][y]$
把 $x,y$ 看成二维坐标上的点，则上图的sum可以用二维差分的方式求出

code

#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll pre[105][105]={0};
int suf[105][105]={0};
ll dp[105][105][105]={0};
const ll mod=1e9+7;
int main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    ll n,m;
    cin>>n>>m;
    for(ll i=1;i<=n;i++)
    {
        string s;
        cin>>s;
        for(ll j=1;j<=m;j++)
        {
            pre[i][j]=pre[i][j-1]+(s[j-1]=='X');
        }
    }

    for(ll i=n;i>=1;i--)
    {
        for(ll l=1;l<=m;l++)
        {
            for(ll r=m;r>=l;r--)
            {
                if(pre[i][r]-pre[i][l-1]!=0) continue;
                if(i==n) dp[i][l][r]=1;
                else dp[i][l][r]=suf[l][r];
                dp[i][l][r]%=mod;
            }
        }

        for(int l=1;l<=m;l++)
        {
            for(int r=m;r>=1;r--)
            {
                suf[l][r]=((suf[l-1][r]+suf[l][r+1])%mod-suf[l-1][r+1])%mod+dp[i][l][r];
                suf[l][r]%=mod;
            }
        }
    }

    ll ans=1;
    for(ll i=1;i<=n;i++)
    {
        for(ll l=1;l<=m;l++)
        {
            for(ll r=l;r<=m;r++)
            {
                ans+=dp[i][l][r];
                //printf("(%d,%d,%d)  %d\n",i,l,r,dp[i][l][r]);
                ans%=mod;
            }
        }
    }
    cout<<ans;
    return 0;
}