Codeforces Round #257 (Div. 2) 前四题
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<string> 5 #include<algorithm> 6 #include<cmath> 7 #include<queue> 8 9 using namespace std; 10 11 #define ll long long 12 #define eps 1e-8 13 #define mod 1000007 14 #define mnx 5010 15 16 struct peo{ 17 int v, id; 18 }a[mnx]; 19 int main(){ 20 int n, m; 21 scanf( "%d%d", &n, &m ); 22 queue<peo> que; 23 for( int i = 1; i <= n; i++ ){ 24 scanf( "%d", &a[i].v ); 25 a[i].id = i; 26 que.push(a[i]); 27 } 28 while( que.size() != 1 ){ 29 peo tmp = que.front(); 30 que.pop(); 31 if( tmp.v > m ){ 32 tmp.v -= m; 33 que.push(tmp); 34 } 35 } 36 peo tmp = que.front(); 37 printf( "%d\n", tmp.id ); 38 return 0; 39 }
f[i] = f[i-1] + f[i+1] 即 f[i+1] = f[i] - f[i-1] 则 f[i+2] = f[i+1] - f[i] = f[i] - f[i-1] - f[i] = - f[i-1]，即 f[i+3] = -f[i]
可以得到 f[i+6] = -f[i+3] = f[i] 循环节是6
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<string> 5 #include<algorithm> 6 #include<cmath> 7 #include<queue> 8 9 using namespace std; 10 11 #define ll long long 12 #define eps 1e-8 13 #define mod 1000000007 14 #define mnx 5010 15 16 ll a[mnx]; 17 int main(){ 18 ll x, y, n; 19 cin >> x >> y >> n; 20 a[0] = x, a[1] = y, a[2] = a[1] - a[0]; 21 a[3] = -x, a[4] = -y, a[5] = -a[2]; 22 n = ( n - 1 ) % 6; 23 ll ans = a[n]; 24 while( ans < 0 ){ 25 ans += mod; 26 } 27 cout << ans % mod <<endl; 28 return 0; 29 }
n*m巧克力，横切和纵切总共k刀,求最小矩形的最大面积是多少。。因为如果横切和纵切混着来，巧克力就会越分越多，所以沿着其中一条边切，如果这一条边不能再切了才去切另外一条，这样能够保证k刀下来，最小巧克力的面积达到最大。。沿着两条边都试切一下，取最大值
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<string> 5 #include<algorithm> 6 #include<cmath> 7 #include<queue> 8 9 using namespace std; 10 11 #define ll long long 12 #define eps 1e-8 13 #define mod 1000000007 14 #define mnx 5010 15 16 int main(){ 17 ll n, m, k; 18 cin >> n >> m >> k; 19 ll k1 = k; 20 ll sum = n + m - 2, ans, ans1, sum1 = n * m; 21 if( k > sum ){ 22 puts( "-1" ); 23 } 24 else if( k == sum ){ 25 puts( "1" ); 26 } 27 else{ 28 if( k + 1 <= n ){ 29 ans = n / ( k + 1 ) * m; 30 } 31 if( k + 1 > n ){ 32 k -= n - 1; 33 ans = m / ( k + 1 ); 34 } 35 swap( n, m ); 36 if( k1 + 1 <= n ){ 37 ans1 = n / ( k1 + 1 ) * m; 38 } 39 if( k1 + 1 > n ){ 40 k1 -= n - 1; 41 ans1 = m / ( k1 + 1 ); 42 } 43 cout << max( ans, ans1 ) <<endl; 44 } 45 return 0; 46 }
最短路。。有n个城市，1是首都，m条路( u, v, c1 )，还有k条火车路线( 1, v, c2 )，问你最多能够删除多少条火车路，使得从1到所有城市的最短路不变。。首先以1为起点对m条路进行dij()，然后k条火车路线( 1, v, c2 ),如果c2 >= dis[v] 的话，这条路线就可以删去；但是如果c2 < dis[v]的话，就要讨论一下能不能删除: 假如dis[u] = 100 , dis[v] = 100, 有一条路( u, v, 60 ), 火车路线( 1, u, 10 ), ( 1, v, 80 ),虽然两条火车路线都比dis[]要短，但是( 1, v, 80 )这一条是可以删除的，因为从( 1, u, 10 )这条线加上( u, v, 60 )距离小于( 1, v, 80 )。所以要对c2 < dis[v]的点再dij() 一次
1 #include<iostream> 2 #include<cstring> 3 #include<algorithm> 4 #include<cstdio> 5 #include<string> 6 #include<queue> 7 #include<cmath> 8 9 using namespace std; 10 11 #define mnx 800500 12 #define ll long long 13 #define inf 0x3f3f3f3f 14 #define lson l, m, rt << 1 15 #define rson m+1, r, rt << 1 | 1 16 17 int vv[mnx], first[mnx], nxt[mnx], e, ans; 18 ll cost[mnx], dis[mnx], dis2[mnx]; 19 bool vis[mnx], used[mnx], in[mnx]; 20 int n, m, k; 21 void init(){ 22 memset( first, -1, sizeof(first) ); 23 memset( dis, 0x3f, sizeof(dis) ); 24 memset( vis, 0, sizeof(vis) ); 25 e = ans = 0; 26 } 27 void add( int u, int v, int c ){ 28 vv[e] = v, nxt[e] = first[u]; 29 cost[e] = c, first[u] = e++; 30 } 31 struct edge{ 32 int u; 33 ll d; 34 bool operator < ( const edge & b ) const{ 35 return d > b.d; 36 } 37 }; 38 void dij(){ 39 priority_queue<edge> que; 40 dis[1] = 0; 41 edge tmp; 42 tmp.u = 1, tmp.d = 0; 43 que.push( tmp ); 44 while( !que.empty() ){ 45 tmp = que.top(); que.pop(); 46 int u = tmp.u; 47 if( vis[u] ) continue; 48 vis[u] = 1; 49 for( int i = first[u]; i != -1; i = nxt[i] ){ 50 int v = vv[i]; ll c = cost[i]; 51 if( dis[v] > dis[u] + c ){ 52 dis[v] = dis[u] + c; 53 edge q; 54 q.u = v, q.d = dis[v]; 55 que.push( q ); 56 } 57 } 58 } 59 } 60 void dij2(){ 61 memset( vis, 0, sizeof(vis) ); 62 priority_queue<edge> que; 63 for( int i = 2; i <= n; i++ ){ 64 if( dis2[i] < dis[i] ){ 65 edge q; 66 q.u = i, q.d = dis2[i]; 67 que.push( q ); 68 } 69 } 70 while( !que.empty() ){ 71 edge q = que.top(); que.pop(); 72 int u = q.u; 73 if( vis[u] ) continue; 74 vis[u] = 1; 75 for( int i = first[u]; i != -1; i = nxt[i] ){ 76 int v = vv[i]; ll c = cost[i]; 77 if( dis2[v] >= dis2[u] + c ){ 78 dis2[v] = dis2[u] + c; 79 if( used[v] ){ 80 ans++; 81 used[v] = 0; 82 } 83 edge q2; 84 q2.u = v, q2.d = dis2[v]; 85 que.push( q2 ); 86 } 87 } 88 } 89 } 90 int main(){ 91 init(); 92 scanf( "%d%d%d", &n, &m, &k ); 93 for( int i = 0; i < m; i++ ){ 94 int u, v; ll c; 95 scanf( "%d%d%I64d", &u, &v, &c ); 96 add( u, v, c ); 97 add( v, u, c ); 98 } 99 dij(); 100 memcpy( dis2, dis, sizeof(dis2) ); 101 memset( used, 0, sizeof(used) ); 102 for( int i = 0; i < k; i++ ){ 103 int v; ll c; 104 scanf( "%d%I64d", &v, &c ); 105 if( dis2[v] <= c ) ans++; 106 else{ 107 dis2[v] = c; 108 if( used[v] ) ans++; 109 used[v] = 1; 110 } 111 } 112 dij2(); 113 printf( "%d\n", ans ); 114 return 0; 115 }
