| 10285 - Longest Run on a Snowboard | 4910 | 55.11% | 2036 | 93.76% |
| 10075943 | 10285 | Longest Run on a Snowboard | Accepted | C++ | 0.024 | 2012-05-05 02:12:46 |
滑雪， 变相的最长递减子序列，它不是求一行的最长递增子序列，而是求在一个方阵，可以往上下左右四个方向走的最长递减路线。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int max(int a,int b){return a>b?a:b;}
int map[105][105],dp[105][105];
int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};
class CPoint{
public:
int height;
int x;
int y;
friend bool operator < (const CPoint &a,const CPoint &b){
return a.height < b.height;
}
}road[10005];
int main(int i, int j)
{
// freopen("input.txt","r",stdin);
char name[100];
int N,R,C;
scanf("%d",&N);
while(N--){
scanf("%s %d %d",name,&R,&C);
int numRoad=0;
for(i=0; i<R; ++i){
for(j=0; j<C; ++j){
scanf("%d",&map[i][j]);
road[numRoad].height=map[i][j];
road[numRoad].x=i;
road[numRoad].y=j;
++numRoad;
}
}
sort(road,road+numRoad);
// memset(dp,1,sizeof(dp)); /* 不能用这个，无法初始化为1 */
for(i=0; i<R; ++i){
for(j=0; j<C; ++j)
dp[i][j]=1;
}
int maxRoad = 0;
for(i=0; i<numRoad; ++i){
for(j=0; j<4; ++j){
int dx=road[i].x+dir[j][0];
int dy=road[i].y+dir[j][1];
if(dx>=0 && dx<R && dy>=0 && dy<C && map[dx][dy]>map[road[i].x][road[i].y] && dp[road[i].x][road[i].y]+1>dp[dx][dy]){
dp[dx][dy] = dp[road[i].x][road[i].y]+1;
if(dp[dx][dy]>maxRoad) maxRoad = dp[dx][dy];
}
}
}
printf("%s: %d\n",name,maxRoad);
}
return 0;
}这题WA了多次，原因在于不懂的打印出 最长公共子序列。 网上查了下，学习了两种打印方式：最长公共子序列问题，打印路径总结了两个方法，一个是递归指向，就像前面做的01背包问题如：
for(int i = 1; i < n; i ++)
for(int j = 0; j < n-i; j ++)
for(int k = j; k < i+j; k ++)
{
if(k == j)
{
f[j][i+j] = f[j][k]+f[k+1][i+j]+A[j][0]*A[k][1]*A[i+j][1];
part1[j][i+j] = k;//赋值
}
if(f[j][k]+f[k+1][i+j]+A[j][0]*A[k][1]*A[j+i][1] < f[j][i+j])
{
f[j][i+j] = f[j][k]+f[k+1][i+j]+A[j][0]*A[k][1]*A[j+i][1];
part1[j][i+j] = k;//赋值
}
}void printpath(int a, int b)//输出
{
if(a == b)
{
printf("A%d", a + 1);
return ;
}
printf("(");
printpath(a, part1[a]);
printf(" x ");
printpath(part1[a][b] + 1, b);
printf(")");
}例如：
for(int i = 1; i <= cdnum; i ++)
for(int j = 0; j <= tap; j ++)
{
f[i][j] = f[i-1][j];
p[i][j] = j;//赋值
if(j>=num[i]&&f[i-1][j-num[i]]+num[i] > f[i][j])
{
f[i][j] = f[i-1][j-num[i]]+num[i]; p[i][j] = j-num[i];//赋值
}
}void print(int i, int j)//打印路径
{
if(i == 0)
return;
print(i - 1, p[i][j]);
if(p[i][j] < j)
printf("%d ", num[i]);
}这第二种就是今天用到的 赋值遍历打印路径；
for(int i = 1; i <= m; i ++)
for(int j = 1; j <= n; j ++)
if(!strcmp(a[i], b[j]))
{
f[i][j] = f[i-1][j-1] + 1;
p[i][j] = 1;//赋值
}
else if(f[i-1][j]>f[i][j-1])
{
f[i][j] = f[i-1][j];
p[i][j] = 0;//赋值
}
else
{
f[i][j] = f[i][j-1];
p[i][j] = -1;//赋值
}void print(int i , int j)//输出{ if(!i || !j) return ; if(p[i][j] == 1) { print(i-1,j-1); if(flag) printf(" "); else flag = 1; printf("%s", a[i]); } else if(p[i][j] == 0) print(i-1,j); else print(i,j-1);}/*UVa 531 Compromise最长公共子序列打印 */#include<iostream>#include<string>#include<cstring>#include<cstdio>using namespace std;string text1[105],text2[105];int dp[105][105],p[105][105];bool flag;void print(int i,int j){ if(!i || !j) return ; if(p[i][j]==1){ print(i-1,j-1); if(flag) cout<<" "; else flag=true; cout << text1[i]; } else if(p[i][j]==0){ print(i-1,j); } else print(i,j-1);}int main(int i,int j){ freopen("input.txt","r",stdin); int len1,len2; while(cin>>text1[1]){ len1 = 2; len2 = 1; while(cin >> text1[len1]){ if(text1[len1]=="#") break; ++len1; } while(cin >> text2[len2]){ if(text2[len2]=="#") break; ++len2; } --len1; --len2; memset(dp,0,sizeof(dp)); for(i=1; i<=len1; ++i){ for(j=1; j<=len2; ++j){ if(text1[i]==text2[j]){ dp[i][j] = dp[i-1][j-1]+1; p[i][j] = 1; } else{ if(dp[i-1][j]>dp[i][j-1]){ dp[i][j] = dp[i-1][j]; p[i][j] = 0; } else{ dp[i][j] = dp[i][j-1]; p[i][j] = -1; } } } } flag=false; print(len1,len2); printf("\n"); } return 0;}624 - CD (01背包打印路径)
#include<cstdio>#include<cstring>#define MAXN 10000int track[MAXN];int dp[22][MAXN];int path[22][MAXN];int max(int a,int b){return a>b?a:b;}void print(int i, int j)//打印路径{ if(i == 0) return; print(i-1, path[i][j]); if(path[i][j] < j) printf("%d ",track[i]);}int main(){ //freopen("input.txt","r",stdin); int N,t,i,j; while(scanf("%d %d",&N,&t)!=EOF){ for(i=1; i<=t; ++i) scanf("%d",&track[i]); memset(dp,0,sizeof(dp)); memset(path,0,sizeof(path)); for(i=1; i<=t; ++i){ for(j=0; j<=N; ++j){ dp[i][j] = dp[i-1][j]; path[i][j] = j; if(j>=track[i]&&dp[i-1][j-track[i]]+track[i]>dp[i][j] &&dp[i-1][j-track[i]]+track[i]<=N){ dp[i][j]=dp[i-1][j-track[i]]+track[i]; path[i][j]=j-track[i]; } } } print(t,N); printf("sum:%d\n",dp[t]); } return 0;}[b]—— 生命的意义，在于赋予它意义。
原创 http://blog.csdn.net/shuangde800 ， By D_Double
