差分约束 HDU - 1384 HDU - 3592 HDU - 1531 HDU - 3666
Intervals
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5488 Accepted Submission(s): 1999
Write a program that:
> reads the number of intervals, their endpoints and integers c1, ..., cn from the standard input,
> computes the minimal size of a set Z of integers which has at least ci common elements with interval [ai, bi], for each i = 1, 2, ..., n,
> writes the answer to the standard output
Process to the end of file.
3 7 3
8 10 3
6 8 1
1 3 1
10 11 1
#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define rb(a) scanf("%lf", &a)
#define rf(a) scanf("%f", &a)
#define pd(a) printf("%d\n", a)
#define plld(a) printf("%lld\n", a)
#define pc(a) printf("%c\n", a)
#define ps(a) printf("%s\n", a)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _ ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = , INF = 0x7fffffff;
int head[maxn], nex[maxn << ], d[maxn], vis[maxn];
int cnt, n, ans[maxn];
struct node
{
int u, v, w;
}Node[maxn << ]; void add(int u, int v, int w)
{
Node[cnt].u = u;
Node[cnt].v = v;
Node[cnt].w = w;
nex[cnt] = head[u];
head[u] = cnt++;
} bool spfa(int s)
{
mem(d, -0x3f);
queue<int> Q;
Q.push(s);
vis[s] = ;
d[s] = ;
while(!Q.empty())
{
int u = Q.front(); Q.pop();
vis[u] = ;
for(int i = head[u]; i != -; i = nex[i])
{
int v = Node[i].v;
if(d[v] < d[u] + Node[i].w)
{
d[v] = d[u] + Node[i].w;
if(!vis[v])
{
Q.push(v);
vis[v] = ;
if(++ans[v] > n) return ;
}
}
}
}
return ;
} int main()
{ while(~scanf("%d", &n))
{
mem(head, -);
mem(vis, );
mem(ans, );
cnt = ;
int u, v, w;
int mx = -INF, mi = INF;
for(int i = ; i <= n; i++)
{
rd(u), rd(v), rd(w);
u++, v++;
mx = max(mx, v);
mi = min(mi, u);
add(u - , v, w);
}
for(int i = mi; i <= mx; i++)
{
// add(mi, i, 0);
add(i - , i, );
add(i, i - , -);
} spfa(mi - ); pd(d[mx]); } return ;
}
World Exhibition
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2440 Accepted Submission(s): 1171
There is something interesting. Some like each other and want to be within a certain distance of each other in line. Some really dislike each other and want to be separated by at least a certain distance. A list of X (1 <= X <= 10,000) constraints describes which person like each other and the maximum distance by which they may be separated; a subsequent list of Y constraints (1 <= Y <= 10,000) tells which person dislike each other and the minimum distance by which they must be separated.
Your job is to compute, if possible, the maximum possible distance between person 1 and person N that satisfies the distance constraints.
The next line: Three space-separated integers: N, X, and Y.
The next X lines: Each line contains three space-separated positive integers: A, B, and C, with 1 <= A < B <= N. Person A and B must be at most C (1 <= C <= 1,000,000) apart.
The next Y lines: Each line contains three space-separated positive integers: A, B, and C, with 1 <= A < B <= C. Person A and B must be at least C (1 <= C <= 1,000,000) apart.
4 2 1
1 3 8
2 4 15
2 3 4
#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _ ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = , INF = 0x7fffffff;
int head[maxn], ans[maxn], vis[maxn];
int n, m, cnt, s;
int d[maxn]; struct node
{
int v, next;
int w;
}Node[maxn*]; void add(int u, int v, int w)
{
Node[cnt].v = v;
Node[cnt].w = w;
Node[cnt].next = head[u];
head[u] = cnt++;
} int spfa()
{
// mem(vis, 0);
for(int i = ; i < maxn; i++) d[i] = INF;
queue<int> Q;
mem(ans, );
Q.push(s);
vis[s] = ;
d[s] = ;
// mem(vis, 0);
// for(int i=1; i<=n; i++)
// {
// Q.push(i);
// d[i] = INF;
// vis[i] = 1;
// }
while(!Q.empty())
{
int u = Q.front(); Q.pop();
vis[u] = ;
for(int i=head[u]; i!=-; i=Node[i].next)
{
node e = Node[i];
if(d[e.v] > d[u] + e.w)
{
d[e.v] = d[u] + e.w;
if(!vis[e.v])
{
Q.push(e.v);
vis[e.v] = ;
if(++ans[e.v] > n) return ;
}
}
}
}
return ;
} void init()
{
mem(head, -);
cnt = ;
} bool check(int x)
{
bool flag = ;
for(int i = ; i < cnt; i++)
Node[i].w -= x;
if(spfa())
flag = ;
for(int i = ; i < cnt; i++)
Node[i].w += x;
return flag;
} int main()
{
int x;
int T;
scanf("%d", &T);
while(T--)
{
scanf("%d%d%d", &n, &m, &x);
init();
int u, v, w, l = , r = ;
s = ;
rap(i, , m)
{
scanf("%d%d%d", &u, &v, &w);
if(u > v) swap(u, v);
add(u, v, w);
}
rap(i, , x)
{
scanf("%d%d%d", &u, &v, &w);
if(u > v) swap(u, v);
add(v, u, -w);
}
if(spfa()) printf("-1\n");
else if(d[n] == INF)
printf("-2\n");
else
printf("%d\n", d[n]); } return ;
}
King
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2348 Accepted Submission(s): 1052
Unfortunately, as it used to happen in royal families, the son was a little retarded. After many years of study he was able just to add integer numbers and to compare whether the result is greater or less than a given integer number. In addition, the numbers had to be written in a sequence and he was able to sum just continuous subsequences of the sequence.
The old king was very unhappy of his son. But he was ready to make everything to enable his son to govern the kingdom after his death. With regards to his son's skills he decided that every problem the king had to decide about had to be presented in a form of a finite sequence of integer numbers and the decision about it would be done by stating an integer constraint (i.e. an upper or lower limit) for the sum of that sequence. In this way there was at least some hope that his son would be able to make some decisions.
After the old king died, the young king began to reign. But very soon, a lot of people became very unsatisfied with his decisions and decided to dethrone him. They tried to do it by proving that his decisions were wrong.
Therefore some conspirators presented to the young king a set of problems that he had to decide about. The set of problems was in the form of subsequences Si = {aSi, aSi+1, ..., aSi+ni} of a sequence S = {a1, a2, ..., an}. The king thought a minute and then decided, i.e. he set for the sum aSi + aSi+1 + ... + aSi+ni of each subsequence Si an integer constraint ki (i.e. aSi + aSi+1 + ... + aSi+ni < ki or aSi + aSi+1 + ... + aSi+ni > ki resp.) and declared these constraints as his decisions.
After a while he realized that some of his decisions were wrong. He could not revoke the declared constraints but trying to save himself he decided to fake the sequence that he was given. He ordered to his advisors to find such a sequence S that would satisfy the constraints he set. Help the advisors of the king and write a program that decides whether such a sequence exists or not.
1 2 gt 0
2 2 lt 2
1 2
1 0 gt 0
1 0 lt 0
0
successful conspiracy
#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _ ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = , INF = 0x7fffffff;
int head[maxn], ans[maxn], vis[maxn];
int n, m, cnt, s;
int d[maxn]; struct node
{
int v, next;
int w;
}Node[maxn*]; void add(int u, int v, int w)
{
Node[cnt].v = v;
Node[cnt].w = w;
Node[cnt].next = head[u];
head[u] = cnt++;
} int spfa()
{
mem(vis, );
mem(d, 0x3f);
queue<int> Q;
mem(ans, ); for(int i=; i<=n; i++)
{
Q.push(i);
d[i] = ;
vis[i] = ;
}
while(!Q.empty())
{
int u = Q.front(); Q.pop();
vis[u] = ;
for(int i=head[u]; i!=-; i=Node[i].next)
{
node e = Node[i];
if(d[e.v] > d[u] + e.w)
{
d[e.v] = d[u] + e.w;
if(!vis[e.v])
{
Q.push(e.v);
vis[e.v] = ;
if(++ans[e.v] > n) return ;
}
}
}
}
return ;
} void init()
{
mem(head, -);
cnt = ;
} bool check(int x)
{
bool flag = ;
for(int i = ; i < cnt; i++)
Node[i].w -= x;
if(spfa())
flag = ;
for(int i = ; i < cnt; i++)
Node[i].w += x;
return flag;
} int main()
{
int x;
while(scanf("%d", &n) && n)
{
scanf("%d", &m);
init();
int u, v, w, l = , r = ;
s = n + ;
char str[];
rap(i, , m)
{
scanf("%d%d%s%d", &u, &v, str, &w);
if(str[] == 'g')
add(u - , u + v, - - w);
else
add(u + v, u - , w - );
} bool flag = spfa();
if(flag)
cout << "successful conspiracy" << endl;
else
cout << "lamentable kingdom" << endl; } return ;
}
THE MATRIX PROBLEM
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9774 Accepted Submission(s): 2508
Each case includes two parts, in part 1, there are four integers in one line, N,M,L,U, indicating the matrix has N rows and M columns, L is the lowerbound and U is the upperbound (1<=N、M<=400,1<=L<=U<=10000). In part 2, there are N lines, each line includes M integers, and they are the elements of the matrix.
2 3 4
8 2 6
5 2 9
可以得出要求是 L <= num[i][j] * a[i] / b[j] <= U
可以转换一下变成
log(L / num[i][j]) <= log(a[i]) - log(b[i]) <= log(U / num[i][j])
#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define rb(a) scanf("%lf", &a)
#define rf(a) scanf("%f", &a)
#define pd(a) printf("%d\n", a)
#define plld(a) printf("%lld\n", a)
#define pc(a) printf("%c\n", a)
#define ps(a) printf("%s\n", a)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _ ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = , INF = 0x7fffffff;
int n, m, s;
int head[maxn], cnt, nex[maxn << ], vis[maxn];
int ans[maxn];
double d[maxn];
struct node
{
int u, v;
double w;
}Node[maxn << ]; void add(int u, int v, double w)
{
Node[cnt].u = u;
Node[cnt].v = v;
Node[cnt].w = w;
nex[cnt] = head[u];
head[u] = cnt++;
} int spfa()
{
for(int i = ; i < maxn; i++) d[i] = INF;
mem(ans, );
deque<int> Q;
Q.push_front(s);
mem(vis, );
vis[s] = ;
d[s] = ;
while(!Q.empty())
{
int u = Q.front(); Q.pop_front();
vis[u] = ;
for(int i = head[u]; i != -; i = nex[i])
{
int v = Node[i].v;
if(d[v] > d[u] + Node[i].w)
{
d[v] = d[u] + Node[i].w;
if(!vis[v])
{
if(Q.empty()) Q.push_front(v);
else if(d[v] < d[Q.front()]) Q.push_front(v);
else Q.push_back(v);
vis[v] = ;
if(++ans[v] > n) return ;
}
}
}
}
return ;
} int main()
{
int l, r;
while(scanf("%d%d%d%d", &n, &m, &l, &r) != EOF)
{
mem(head, -);
cnt = ;
int u, v, x;
s = ;
for(int i = ; i <= n; i++)
for(int j = ; j <= m; j ++)
{
rd(x);
add(i, n + j, -log(l / (double) x));
add(n + j, i, log(r / (double) x));
} if(spfa())
cout << "NO" << endl;
else cout << "YES" << endl; } return ;
}
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