标签：
嵌套矩形问题(最长路及其字典序)
有n个举行，选出尽量多的矩阵排成一排，使得除了最后一个之外，每一个矩形可以嵌套在下一个矩形内,并且打印
#include <iostream> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <cstdlib> #include <stack> #include <cctype> #include <string> #include <queue> #include <map> #include <set> using namespace std; const int INF = 0xffffff; const double esp = 10e-8; const double Pi = 4 * atan(1.0); const int maxn = 100+ 10; const long long mod = 1000000007; const int dr[] = {1,0,-1,0,-1,1,-1,1}; const int dc[] = {0,1,0,-1,1,-1,-1,1}; typedef long long LL; LL gac(LL a,LL b){ return b?gac(b,a%b):a; } int n; bool graph[maxn][maxn]; struct G{ int a,b; }; G g[maxn]; int d[maxn][maxn]; bool judge(int x,int y){ if(g[x].a < g[y].a && g[x].b < g[y].b) return 1; if(g[x].a < g[y].b && g[x].b < g[y].a) return 1; return 0; } int dp(int i,int tt){ if(d[tt][i] > 0) return d[tt][i]; d[tt][i] = 1; for(int j = 1;j <= n;j++){ if(graph[i][j]){ d[tt][i] = max(d[tt][i],dp(j,tt)+1); } } return d[tt][i]; } void print_ans(int i,int tt){ printf("%d ",i); for(int j = 1;j <= n;j++){ if(graph[i][j] && d[tt][i] == d[tt][j] + 1){ print_ans(j,tt); break; } } } int main() { #ifndef ONLINE_JUDGE freopen("input.in","r",stdin); // freopen("output.txt","w",stdout); #endif int t; scanf("%d",&t); while(t--){ scanf("%d",&n); for(int i = 1;i <= n;i++){ getchar(); char ch; scanf("%c%d%d",&ch,&g[i].a,&g[i].b); //int fin = -1; for(int j = 1;j < i;j++){ if(judge(i,j)){ graph[i][j] = 1; } if(judge(j,i)){ graph[j][i] = 1; } } } int ans = -1; int m; memset(d,0,sizeof(d)); for(int i = 1;i <= n;i++){ int tmp = dp(i,i); if(tmp > ans){ m = i; ans = tmp; } } printf("%d\n",ans); print_ans(m,m); } return 0; }
硬币问题
有n种硬币，面值分别为v1，v2，v3……给定非负整数s问选用多少个硬币使面值恰好等于s？求出硬币数目最大值和最小值……
#include <iostream> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <cstdlib> #include <stack> #include <cctype> #include <string> #include <queue> #include <map> #include <set> using namespace std; const int INF = 0xffffff; const double esp = 10e-8; const double Pi = 4 * atan(1.0); const int maxn = 100+ 10; const long long mod = 1000000007; const int dr[] = {1,0,-1,0,-1,1,-1,1}; const int dc[] = {0,1,0,-1,1,-1,-1,1}; typedef long long LL; LL gac(LL a,LL b){ return b?gac(b,a%b):a; } int V[maxn]; int minv[maxn]; int maxv[maxn]; int d[maxn]; int n; int dp(int i){ if(d[i] != -1) return d[i]; d[i] = -INF; for(int j = 1;j <= n;j++){ if(i >= V[j]){ d[i] = max(d[i],dp(i - V[j])+1); } } return d[i]; } void print_ans(int * a,int s){ for(int i = 1;i <= n;i++){ if(s >= V[i] && a[s] == a[s -V[i] ]+1){ printf("%d ",V[i]); print_ans(a,s-V[i]); break; } } } void print_Ans(int *a,int s){ while(s){ printf("%d ",a[s]); s -= a[s]; } } int main() { #ifndef ONLINE_JUDGE freopen("input.in","r",stdin); // freopen("output.txt","w",stdout); #endif while(~scanf("%d",&n)){ int s; scanf("%d",&s); for(int i = 1;i <= n;i++){ scanf("%d",&V[i]); } d[0] = 0; for(int i = 1;i <= s;i++) d[i] = -1; minv[0] = maxv[0] = 0; for(int i = 1;i <= s;i++){ minv[i] = INF; maxv[i] = -INF; } int min_coin[maxn]; int max_coin[maxn]; for(int i = 1;i <= s;i++){ for(int j = 1;j <= n;j++){ if(i >= V[j]){ if(minv[i] > minv[i - V[j]]+1){ minv[i] = min(minv[i],minv[i-V[j] ] + 1); min_coin[i] = V[j]; } if(maxv[i] < maxv[i-V[j] ]+1){ maxv[i] = max(maxv[i],maxv[i -V[j] ]+1); max_coin[i] = V[j]; } minv[i] = min(minv[i],minv[i-V[j] ] + 1); maxv[i] = max(maxv[i],maxv[i -V[j] ]+1); } } } print_Ans(min_coin,s); printf("\n"); print_Ans(max_coin,s); printf("\n"); printf("%d %d\n",minv[s],maxv[s]); print_ans(minv,s); printf("\n"); print_ans(maxv,s); printf("\n"); int ans = dp(s) ; if(ans < -1){ printf("no problem\n"); } else{ printf("%d\n",ans); } } return 0; }
标签：
原文地址：http://www.cnblogs.com/hanbinggan/p/4333004.html
