Codeforces Round #142 (Div. 2)

　　A.贪心，每次都选剩下的里面x最小的。

//Oct 1, 2012 7:42:03 PM 	bigoceanlhy 	230A - Dragons 	GNU C++ 	Accepted 	46 ms 	0 KB
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN = 1000+5;
int N, s, x[MAXN], y[MAXN], r[MAXN];
bool cmp(const int a, const int b)
{
return x[a] < x[b];
}
int main()
{
scanf("%d%d", &s, &N);
for (int i = 0; i < N; i++)
{
scanf("%d%d", &x[i], &y[i]);
r[i] = i;
}
sort(r, r+N, cmp);
bool flag = 1;
for (int i = 0; i < N && flag; i++)
{
if (s > x[r[i]])
s += y[r[i]];
else
flag = 0;
}
printf("%s\n", flag ? "YES" : "NO");
return 0;
}

　　B.满足条件的数是质数的平方。

//Oct 1, 2012 8:03:30 PM 	bigoceanlhy 	230B - T-primes 	GNU C++ 	Accepted 	296 ms 	1400 KB
#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
const int MAXN = 100000+5, MAXM = 1000000+5;
const double eps = 1e-8;
int N;
long long x;
bool is[MAXM]; int prm[MAXN];
int getprm(int n){
int i, j, k = 0;
int s, e = (int)(sqrt(0.0 + n) + 1);
memset(is, 1, sizeof(is));
prm[k++] = 2; is[0] = is[1] = 0;
for (i = 4; i < n; i += 2) is[i] = 0;
for (i = 3; i < e; i += 2) if (is[i]) {
prm[k++] = i;
for (s = i * 2, j = i * i; j < n; j += s)
is[j] = 0;
}
for ( ; i < n; i += 2) if (is[i]) prm[k++] = i;
return k;
}
int main()
{
int c = getprm(1000003);
scanf("%d", &N);
for (int i = 0; i < N; i++)
{
cin>>x;
long long t = sqrt(x+0.0)+eps;
if (t*t == x && is[(int)t])
printf("YES\n");
else
printf("NO\n");
}
return 0;
}

　　C.DP，d[i][j]表示前i层的第j列都为1的最小代价。

//Oct 1, 2012 10:49:36 PM 	bigoceanlhy 	229A - Shifts 	GNU C++ 	Accepted 	46 ms 	5100 KB
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN = 100+5, MAXM = 10000+5;
const int INF = 0x3f3f3f3f;
int N, M, d[MAXN][MAXM];
char mat[MAXN][MAXM];
int main()
{
scanf("%d%d", &N, &M);
for (int i = 1; i <= N; i++)
scanf("%s", mat[i]+1);
fill(d[0]+1, d[0]+1+M, 0);
bool flag = 1;
for (int i = 1; i <= N; i++)
{
flag = 0;
for (int j = 1; j <= M; j++)
{
d[i][j] = INF;
if (mat[i][j] == '1')
flag = 1;
}
if (!flag)
break;
int nst;
for (int j = 1; j <= M; j++)
if (mat[i][j] == '1')
nst = j;
nst -= M;
for (int j = 1; j <= M; j++)
{
if (mat[i][j] == '1')
nst = j;
d[i][j] = min(d[i][j], d[i-1][j]+j-nst);
}
for (int j = M; j >= 1; j--)
if (mat[i][j] == '1')
nst = j;
nst += M;
for (int j = M; j >= 1; j--)
{
if (mat[i][j] == '1')
nst = j;
d[i][j] = min(d[i][j], d[i-1][j]+nst-j);
}
}
int ans = INF;
for (int j = 1; j <= M; j++)
ans = min(ans, d
[j]);
if (flag)
printf("%d\n", ans);
else
printf("-1\n");
return 0;
}

　　D.大体思路就是直接最短路，松弛的时候加上等待的时间，要先把每个点有冲突的时刻的等待时间预处理出来。

//Oct 2, 2012 10:30:39 AM 	bigoceanlhy 	229B - Planets 	GNU C++ 	Accepted 	296 ms 	13000 KB
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<map>
using namespace std;
const int MAXN = 100000+5, MAXM = 200000+5;
const int INF = 0x3f3f3f3f;
int N, M, a, b, c, k, t[MAXN];
int e, head[MAXN], next[MAXM], v[MAXM];
int w[MAXM], d[MAXN];
bool inq[MAXN];
queue<int> Q;
map<int, int> wait[MAXN];
void addedge(int x, int y, int z)
{
v[e] = y; w[e] = z;
next[e] = head[x]; head[x] = e++;
}
void spfa(int s)
{
for (int u = 1; u <= N; u++)
d[u] = (u == s ? 0 : INF);
Q.push(s);
while (!Q.empty())
{
int u = Q.front(); Q.pop();
inq[u] = 0;
for (int i = head[u]; i != -1; i = next[i])
{
int x = d[u]+wait[u][d[u]]+w[i];
if (x < d[v[i]])
{
d[v[i]] = x;
if (!inq[v[i]])
{
Q.push(v[i]);
inq[v[i]] = 1;
}
}
}
}
}
int main()
{
memset(head, -1, sizeof(head));
scanf("%d%d", &N, &M);
for (int i = 1; i <= M; i++)
{
scanf("%d%d%d", &a, &b, &c);
addedge(a, b, c);
addedge(b, a, c);
}
for (int i = 1; i <= N; i++)
{
scanf("%d", &k);
for (int j = 1; j <= k; j++)
scanf("%d", &t[j]);
int last = 1;
for (int j = k; j >= 1; j--)
{
if (j < k && t[j]+1 == t[j+1])
wait[i][t[j]] = ++last;
else
wait[i][t[j]] = (last = 1);
}
}
spfa(1);
printf("%d\n", d
< INF ? d
: -1);
return 0;
}