Missile
Time Limit: 2 Seconds
Memory Limit: 65536 KB
You control N missile launching towers. Every tower has enough missiles, but for each tower only one missile can be launch at the same time. Before the launching, every missile needT1 seconds to leave the tower. Assume that all
the missiles have the same speedV, and it would fly along the shortest path to the target. You can just consider the horizontal distance and ignore the height. You can consider the time equal to distance / V (minutes). The missile can immediately
destroy the target when it reached. Besides, for the same tower, after launching a missile, it needT2 minutes to prepare for the next one.
Now, give you the coordinate position of N missile launching towers andM targets,T1,
T2 and V, you should find the minimum minutes to destroy all the targets.
Input
The input will consist of about 10 cases. The first line of each case contains five positive integer numbersN,M,
T1, T2 and V, decribed as above. The nextM lines contains two integer numbers indicating the coordinate ofM targets. The continueingN lines contains two integer numbers indicating the
coordinate ofN towers.
To all the cases, 1 ≤ N ≤ 50, 1 ≤ M ≤ 50
The absolute value of all the coordinates will not exceed 10000, T1,T2,V will not exceed 2000.
Output
For each case, the output is only one line containing only one real number with six digits precision (after a decimal point) indicating the minimum minutes to destroy all the targets.
Sample Input
3 3 30 20 1 0 0 0 50 50 0 50 50 0 1000 1000 0
Sample Output
91.500000
题意：用N个导弹发射塔攻击M个目标。每个导弹发射塔只能同时为一颗导弹服务，发射一颗导弹后需要T1（这里用的是秒）的时间才能离开当前的导弹发射塔，一颗导弹从发射到击中目标的时间与目标到发射塔的距离有关（直线距离），每颗导弹发射完成之后发射塔需要T2的时间准备下一个。现在给出N个导弹发射塔和M个目标的位置坐标以及T1，T2，V，问用这N个导弹发射塔最少需要多少时间可以击毁所有M个目标。
具体实现
一：对每一个导弹发射器，它击中一个目标共有M种情况：分别为在该发射塔第一次发射、第二次发射、第三次发射...一直到第M次发射。因此我们可以把每一个导弹发射塔拆分成M个发射塔，它们与同一个目标的距离是一样的，唯一不同是T1、T2的花销占时不一样。
二：这样的话我们就得到了N*M个点发射器到 M个目标的映射，所表示的关系是当前发射塔击中目标的耗时。
三：设立超级源点0，超级汇点 N * M + M + 1。
四：超级源点到每个目标引一条容量为1的边，每个发射塔（拆分后的）到超级汇点引一条容量为1的边，按若小于当前查询时间的关系（同匹配建边）建边容量为1。
然后跑最大流，判断最大流是否为M。 接着二分查找。
以上解析来自宇哥哥的博客：笑着走完自己的路
笑着走完自己的
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
#define INF 0x3f3f3f3f
#define maxn 5000
#define maxm 500000
using namespace std;
int head[maxn], cur[maxn], cnt;
int dist[maxn], vis[maxn];
struct node {
int u, v, cap, flow, next;
};
struct NODE{
double x, y;
};
node edge[maxm];
NODE tower[60];
NODE target[60];
int N, M;
double T1, T2, V;
double map[60][60];
double Time[3000][60];
double change(NODE a, NODE b){
return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
void init(){
cnt = 0;
memset(head, -1, sizeof(head));
}
void add(int u, int v, int w){
node E1 = {u, v, w, 0, head[u]};
edge[cnt] = E1;
head[u] = cnt++;
node E2 = {v, u, 0, 0, head[v]};
edge[cnt] = E2;
head[v] = cnt++;
}
void getmap(double max_min){
int i, j;
for(i = 1; i <= N * M; ++i)
add(0, i, 1);
for(i = N * M + 1; i <= N * M + M; ++i)
add(i, N * M + M + 1, 1);
for(i = 1; i <= N * M; ++i)
for(j = 1; j <= M; ++j)
if(Time[i][j] <= max_min)
add(i, j + N * M, 1);
}
bool BFS(int st, int ed){
queue<int>q;
memset(vis, 0, sizeof(vis));
memset(dist, -1, sizeof(dist));
q.push(st);
vis[st] = 1;
dist[st] = 0;
while(!q.empty()){
int u = q.front();
q.pop();
for(int i = head[u]; i != -1; i = edge[i].next){
node E = edge[i];
if(!vis[E.v] && E.cap > E.flow){
vis[E.v] = 1;
dist[E.v] = dist[u] + 1;
if(E.v == ed) return true;
q.push(E.v);
}
}
}
return false;
}
int DFS(int x, int ed, int a){
if(a == 0 || x == ed)
return a;
int flow = 0, f;
for(int &i = cur[x]; i != -1; i =edge[i].next){
node &E = edge[i];
if(dist[E.v] == dist[x] + 1 && (f = DFS(E.v, ed, min(a, E.cap - E.flow))) > 0){
E.flow += f;
edge[i ^ 1].flow -= f;
flow += f;
a -= f;
if(a == 0) break;
}
}
return flow;
}
int maxflow (int st, int ed){
int flowsum = 0;
while(BFS(st, ed)){
memcpy(cur, head, sizeof(head));
flowsum += DFS(st, ed, INF);
}
return flowsum;
}
int main (){
while(scanf("%d%d%lf%lf%lf", &N, &M, &T1, &T2, &V) != EOF){
T1 = T1 / 60;
int i, j, k;
for(j = 1; j <= M; ++j)
scanf("%lf%lf", &target[j].x, &target[j].y);
for(i = 1; i <= N; ++i)
scanf("%lf%lf", &tower[i].x, &tower[i].y);
for(i = 1; i <= N; ++i)
for(j = 1; j <= M; ++j)
map[i][j] = change(tower[i], target[j]);
for(i = 1; i <= N; ++i)
for(k = 1; k <= M; ++k){
for(j = 1; j <= M; ++j)
Time[(i - 1) * M + k][j] = T1 * k + T2 * (k - 1) + map[i][j] / V;
}
double l = 0.0, r = 200000000000.0, mid;
while(r - l >= 1e-8){
mid = (l + r) / 2;
init();
getmap(mid);
if(maxflow(0, N * M + M + 1) >= M)
r = mid;
else
l = mid;
}
printf("%.6lf\n", r);
}
return 0;
}
版权声明：本文为博主原创文章，未经博主允许不得转载。