【Solution】
接上一篇,在處理有向無環(huán)圖的最長(zhǎng)鏈問題的時(shí)候,可以在做拓?fù)渑判虻耐瑫r(shí),一邊做DP;
設(shè)f[i]表示第i個(gè)方塊作為最上面的最高值;
f[y]=max(f[y],f[x]+h[y]);(x?>y)∈E
這樣可以保證,按階段進(jìn)行DP,每次在獲取f[x]的時(shí)候,你可以保證f[x]已經(jīng)獲得了;
最后取max(f[1..n])
【Code】

#include <bits/stdc++.h> using namespace std; #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define LL long long #define rep1(i,a,b) for (int i = a;i <= b;i++) #define rep2(i,a,b) for (int i = a;i >= b;i--) #define mp make_pair #define pb push_back #define fi first #define se second #define ms(x,y) memset(x,y,sizeof x) #define Open() freopen("D:\\rush.txt","r",stdin) #define Close() ios::sync_with_stdio(0)typedef pair<int,int> pii; typedef pair<LL,LL> pll;const int dx[9] = {0,1,-1,0,0,-1,-1,1,1}; const int dy[9] = {0,0,0,-1,1,-1,1,-1,1}; const double pi = acos(-1.0); const int N = 30;struct abc{LL c,k,g; };int n,b[4],nn,du[N*3+100]; LL dp[N*3+100]; abc a[N*3+100]; vector <int> G[N*3+100]; queue <int> dl;int main() {//Open();int kk = 0;while (~scanf("%d",&n) && n){kk++;ms(dp,-1);nn = 0;ms(du,0);rep1(i,1,N*3) G[i].clear();rep1(i,1,n){rep1(j,1,3)scanf("%d",&b[j]);sort(b+1,b+1+3);rep1(j,1,3){nn++;rep2(k,3,1)if (k!=j){a[nn].c = b[k];break;}rep1(k,1,3)if (k!=j){a[nn].k = b[k];break;}a[nn].g = b[j];}}n = nn;rep1(i,1,n)rep1(j,1,n)if (a[i].c > a[j].c && a[i].k > a[j].k){G[i].pb(j);du[j]++;}while (!dl.empty()) dl.pop();rep1(i,1,n)if (du[i]==0){dl.push(i);dp[i] = a[i].g;du[i] = -1;}while (!dl.empty()){int x = dl.front();dl.pop();int len = G[x].size();rep1(i,0,len-1){int y = G[x][i];if (dp[y]==-1){dp[y] = dp[x] + a[y].g;}elsedp[y] = max(dp[y],dp[x]+a[y].g);du[y]--;if (du[y]==0){dl.push(y);du[y]= -1;}}}LL d = 0;rep1(i,1,n)d = max(d,dp[i]);printf("Case %d: maximum height = ",kk);printf("%lld\n",d);}return 0; }

轉(zhuǎn)載于:https://www.cnblogs.com/AWCXV/p/7626211.html

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