description

神犇航空有K架飛機，為了簡化問題，我們認為每架飛機都是相同的。神犇航空的世界中有N個機場，以0…N-1編號，其中0號為基地機場，每天0時刻起飛機才可以從該機場起飛，并不晚于T時刻回到該機場。一天，神犇航空接到了M個包機請求，每個請求為在s時刻從a機場起飛，在恰好t時刻到達b機場，可以凈獲利c。設計一種方案，使得總收益最大。

Input
第一行，4個正整數N,M,K,T，如題目描述中所述；

以下N行，每行N個整數，描述一個N*N的矩陣t，t-i,j表示從機場i空載飛至機場j，需要時間ti,j；

以下N行, 每行N個整數，描述一個N*N的矩陣f，f-i,j表示從機場i空載飛至機場j，需要費用fi,j；

以下M行，每行5個整數描述一個請求，依次為a,b,s,t,c。

Output
僅一行，一個整數，表示最大收益。

Sample Input
2 1 1 10
0 5
5 0
0 5
5 0
0 1 0 5 10

Sample Output
5

Hint
數據規模及約定
對于10%的測試數據，K=1；
另有20%的測試數據，K=2；
對于全部的測試數據，N,M<=200，K<=10，T<=3000，ti,j<=200，fi,j<=2000，0<=a,b<N，0<=s<=t<=T，0<=c<=10000，ti,i=fi,i=0，ti,j<=ti,k+tk,j，fi,j<=fi,k+fk,j。

solution

將請求拆成兩個點，之間的費用就是凈收益$c$，流量為$1$，一個請求只完成一次
然后再根據時間限制判斷能否將請求的起終點與源點匯點連邊，流量$inf$

從$0$飛到請求起點的時間$\le$請求要求的起飛時間
從請求要求的結束時間?從請求終點飛回$0$點時間$\le T$

但不一定是完成一個請求后立馬飛回$0$點，或許直接飛往下一個請求的起點更優
所以還需要對請求之間進行兩兩建邊

請求完成時刻?飛到下一個請求的起點所有時間$\le$下一個請求的起飛時刻

最大化收益那就跑最大費用最大流

code

#include <queue> #include <cstdio> #include <cstring> using namespace std; #define maxn 205 #define maxm 100005 #define inf 0x3f3f3f3f struct node {int a, b, s, t, c; }Q[maxn]; struct Edge {int nxt, to, w, flow; }edge[maxm << 1]; queue < int > q; int n, m, K, T, st, ed, cnt, Time, cost; int ti[maxn][maxn], w[maxn][maxn]; int dis[maxm], head[maxm], vis[maxm], pre[maxm];void addedge( int u, int v, int w, int flow ) {edge[cnt].nxt = head[u];edge[cnt].to = v;edge[cnt].w = w;edge[cnt].flow = flow;head[u] = cnt ++; }bool spfa() {for( int i = 0;i <= ed;i ++ ) dis[i] = -inf, vis[i] = 0, pre[i] = -1;//memset(dis,-0x3f,sizeof(dis))無法初始化極小值 q.push( st );dis[st] = 0, vis[st] = 1;while( ! q.empty() ) {int u = q.front(); q.pop();vis[u] = 0;for( int i = head[u];~ i;i = edge[i].nxt ) {int v = edge[i].to;if( dis[v] < dis[u] + edge[i].w && edge[i].flow ) {dis[v] = dis[u] + edge[i].w;pre[v] = i;if( ! vis[v] ) {q.push( v );vis[v] = 1;}}}}return dis[ed] != -inf; }void MCMF() {while( spfa() ) {int flow = inf;for( int i = pre[ed];~ i;i = pre[edge[i ^ 1].to] )flow = min( flow, edge[i].flow );for( int i = pre[ed];~ i;i = pre[edge[i ^ 1].to] ) {edge[i].flow -= flow;edge[i ^ 1].flow += flow;cost += edge[i].w * flow;}} }int main() {memset( head, -1, sizeof( head ) );scanf( "%d %d %d %d", &n, &m, &K, &T );for( int i = 0;i < n;i ++ )for( int j = 0;j < n;j ++ )scanf( "%d", &ti[i][j] );for( int i = 0;i < n;i ++ )for( int j = 0;j < n;j ++ )scanf( "%d", &w[i][j] );for( int i = 1;i <= m;i ++ )scanf( "%d %d %d %d %d", &Q[i].a, &Q[i].b, &Q[i].s, &Q[i].t, &Q[i].c );st = 0, ed = ( m << 1 ) + 2;for( int i = 1;i <= m;i ++ ) {addedge( i << 1, i << 1 | 1, Q[i].c, 1 );addedge( i << 1 | 1, i << 1, -Q[i].c, 0 );if( Q[i].t + ti[Q[i].b][0] <= T ) {addedge( i << 1 | 1, ed, -w[Q[i].b][0], inf );addedge( ed, i << 1 | 1, w[Q[i].b][0], 0 );}else continue;if( ti[0][Q[i].a] <= Q[i].s ) {addedge( st + 1, i << 1, -w[0][Q[i].a], inf );addedge( i << 1, st + 1, w[0][Q[i].a], 0 );}for( int j = 1;j <= m;j ++ )if( Q[i].t + ti[Q[i].b][Q[j].a] <= Q[j].s ) {addedge( i << 1 | 1, j << 1, -w[Q[i].b][Q[j].a], inf );addedge( j << 1, i << 1 | 1, w[Q[i].b][Q[j].a], 0 );}}addedge( st, st + 1, 0, K );addedge( st + 1, st, 0, 0 );MCMF();printf( "%d", cost );return 0; }

Dinic費用流版本

#include <queue> #include <cstdio> #include <cstring> using namespace std; #define maxn 205 #define maxm 100005 #define inf 0x3f3f3f3f struct node {int a, b, s, t, c; }Q[maxn]; struct Edge {int nxt, to, w, flow; }edge[maxm << 1]; queue < int > q; int n, m, K, T, st, ed, cnt = 1, Time, cost; int ti[maxn][maxn], w[maxn][maxn]; int dis[maxm], head[maxm], vis[maxm];void addedge( int u, int v, int w, int flow ) {cnt ++;edge[cnt].nxt = head[u];edge[cnt].to = v;edge[cnt].w = w;edge[cnt].flow = flow;head[u] = cnt; }bool spfa() {for( int i = 0;i <= ed;i ++ ) dis[i] = -inf, vis[i] = 0;//memset(dis,-0x3f,sizeof(dis))無法初始化極小值 q.push( st );dis[st] = 0, vis[st] = 1;while( ! q.empty() ) {int u = q.front(); q.pop();vis[u] = 0;for( int i = head[u];i;i = edge[i].nxt ) {int v = edge[i].to;if( dis[v] < dis[u] + edge[i].w && edge[i].flow ) {dis[v] = dis[u] + edge[i].w;if( ! vis[v] ) {q.push( v );vis[v] = 1;}}}}return dis[ed] != -inf; }int dfs( int u, int cap ) {if( u == ed ) return cap;vis[u] = Time;int flow = 0;for( int i = head[u];i;i = edge[i].nxt ) {int v = edge[i].to;if( ( vis[v] != Time || v == ed ) && edge[i].flow && dis[v] == dis[u] + edge[i].w ) {int tmp = dfs( v, min( cap, edge[i].flow ) );if( ! tmp ) continue;edge[i].flow -= tmp;edge[i ^ 1].flow += tmp;flow += tmp;cap -= tmp;cost += tmp * edge[i].w;if( ! cap ) break;}}return flow; }void Dinic() {while( spfa() ) {do {Time ++;dfs( st, inf );} while( vis[ed] == Time );} }int main() {scanf( "%d %d %d %d", &n, &m, &K, &T );for( int i = 0;i < n;i ++ )for( int j = 0;j < n;j ++ )scanf( "%d", &ti[i][j] );for( int i = 0;i < n;i ++ )for( int j = 0;j < n;j ++ )scanf( "%d", &w[i][j] );for( int i = 1;i <= m;i ++ )scanf( "%d %d %d %d %d", &Q[i].a, &Q[i].b, &Q[i].s, &Q[i].t, &Q[i].c );st = 0, ed = ( m << 1 ) + 2;for( int i = 1;i <= m;i ++ ) {addedge( i << 1, i << 1 | 1, Q[i].c, 1 );addedge( i << 1 | 1, i << 1, -Q[i].c, 0 );if( Q[i].t + ti[Q[i].b][0] <= T ) {addedge( i << 1 | 1, ed, -w[Q[i].b][0], inf );addedge( ed, i << 1 | 1, w[Q[i].b][0], 0 );}else continue;if( ti[0][Q[i].a] <= Q[i].s ) {addedge( st + 1, i << 1, -w[0][Q[i].a], inf );addedge( i << 1, st + 1, w[0][Q[i].a], 0 );}for( int j = 1;j <= m;j ++ )if( Q[i].t + ti[Q[i].b][Q[j].a] <= Q[j].s ) {addedge( i << 1 | 1, j << 1, -w[Q[i].b][Q[j].a], inf );addedge( j << 1, i << 1 | 1, w[Q[i].b][Q[j].a], 0 );}}addedge( st, st + 1, 0, K );addedge( st + 1, st, 0, 0 );Dinic();printf( "%d", cost );return 0; } 創作挑戰賽新人創作獎勵來咯，堅持創作打卡瓜分現金大獎

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