河南省第六届ACM程序设计大赛

C:   最舒适的路线 (并查集)

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define INF 20000000
 
struct point
{
    int x, y, v;
};
 
point node[5005];
int fa[505];
 
bool cmp(point a, point b)
{
    return a.v < b.v;//边从小到大
}
int find(int n)
{
    return n == fa[n] ? n : fa[n] = find(fa[n]);
}
int gcd(int a, int b)
{
    return b ? gcd(b, a%b) : a;
}
int main()
{
    int i, j, t, e;
    int start, end, row, edge, maxv, minv;
    double radio;
    cin>>t;
    while(t--)
    {
        cin>>row>>edge;
        for(i = 1; i <= edge; i++)
        {
            cin>>node[i].x>>node[i].y>>node[i].v;
        }
        cin>>start>>end;
        sort(node + 1, node+edge + 1, cmp);
        radio = INF;
        for(e = edge; e > 0; e--) //从最大的边开始，将比它小的边加入
        {
            for(i =1; i <= row; i++)
                fa[i] = i;
            for(i = e; i > 0; i--)
            {
                int a = find(node[i].x);
                int b = find(node[i].y);
                if(a == b)
                    continue;
                else
                    fa[a] = b;
                if(find(start) == find(end))
                    break;//起点和终点在都一个集合里
            }
            if(i == 0)
                break;//如果把边都加进去完了，说明起点和终点不连通
            if(node[e].v/(node[i].v*1.0) < radio)
            {
                radio = node[e].v / (node[i].v * 1.0);
                maxv = node[e].v;
                minv = node[i].v;
            }
        }
        int g = gcd(maxv, minv);
        if(radio == INF)
            printf("IMPOSSIBLE
");
        else if(maxv % minv == 0)
            printf("%d
", maxv / minv);
        else
            printf("%d/%d
", maxv/g, minv/g);
    }
    return 0;
}

D:  探 寻 宝 藏 (多线程dp)

#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
using namespace std;
typedef unsigned long long LL;
 
#define N 110
#define met(a,b) (memset(a,b,sizeof(a)))
#define max4(a,b,c,d) (max(max(a,b),max(c,d)))
 
 
int a[N][N], dp[N*2][N][N];
 
int main()
{
    int n, m, T;
 
    scanf("%d", &T);
 
    while(T--)
    {
        int i, j, k;
 
        met(a, 0);
        met(dp, 0);
 
        scanf("%d%d", &n, &m);
 
        for(i=1; i<=n; i++)
        for(j=1; j<=m; j++)
            scanf("%d", &a[i][j]);
 
        dp[0][1][1] = a[1][1];
        for(k=1; k<=n+m-2; k++)
        {
            for(i=1; i<=n; i++) ///i 代表第一个人所在的行
            for(j=1; j<=n; j++) ///j 代表第二个人所在的行            
            {
                   dp[k][i][j] = max4(dp[k-1][i][j], dp[k-1][i][j-1], dp[k-1][i-1][j], dp[k-1][i-1][j-1]);
                if(i!=j)
                    dp[k][i][j] += a[i][k+2-i] + a[j][k+2-j];
                else
                     dp[k][i][j] += a[i][k+2-i];
            }
        }
 
        printf("%d
", dp[n+m-2][n][n]);
    }
    return 0;
}

G: Adjacent Bit Counts （递推）

#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
using namespace std;
typedef unsigned long long LL;

#define N 110
#define met(a,b) (memset(a,b,sizeof(a)))
#define max4(a,b,c,d) (max(max(a,b),max(c,d)))


LL dp[N][N][3];

///dp[i][j][0] i位数，fun[i]为j的最后一位为0
///dp[i][j][1] i位数，fun[i]为j的最后一位为1
///dp[i][j][1] dp[i][j][0]与dp[i][j][1]之和

int main()
{
    int T, i, j;

    dp[0][0][0] = dp[0][0][2] = 1;
    dp[1][0][0] = dp[1][0][1] = 1;
    dp[1][0][2] = 2;

    for(i=2; i<N; i++)
    {
        for(j=0; j<i; j++)
        {
            dp[i][j][0] = dp[i-1][j][2];
            dp[i][j][1] = dp[i-1][j][0];
            if(j>=1) dp[i][j][1] += dp[i-1][j-1][1];
            dp[i][j][2] = dp[i][j][0] + dp[i][j][1];
        }
    }

    scanf("%d", &T);

    while(T--)
    {
       int n, m;

       scanf("%d%d", &n, &m);

       printf("%lld
", dp[n][m][2]);
    }
    return 0;
}

H:  River Crossing ( 完全背包)

#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
using namespace std;
typedef unsigned long long LL;
#define INF 0x3f3f3f3f
#define N 1100
#define met(a,b) (memset(a,b,sizeof(a)))
#define max4(a,b,c,d) (max(max(a,b),max(c,d)))

int a[N], dp[N];
///dp[i]代表i个羊一起过河需要花费的最短时间

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int n, m, i, j;

        scanf("%d%d", &n, &m);

        for(i=1; i<=n; i++)
            scanf("%d", &a[i]);

        dp[1] = a[1] + m;
        for(i=2; i<=n; i++)
            dp[i] = dp[i-1]+a[i];

        dp[0] = 0;
        for(i=1; i<=n; i++)
        {
            for(j=i; j<=n; j++)
            {
               dp[j] = min(dp[j], dp[j-i]+dp[i]+m);
            }
        }

        printf("%d
", dp[n]);
    }
    return 0;
}