题解 P3324 【[SDOI2015]星际战争】

· · 个人记录

右转请

Solution

重难点依然是建图
考虑攻击力限制条件
一个是攻击装置能够发射的伤害
一个是防御装置能够承受的伤害
建立超级源点超级汇点
源点向武器建边，边权为攻击参数
防御向汇点建边，边权为防御系数
武器与防御之间边权inf即可
考虑二分答案
加入现在拥有一个攻击时间
那么每个武器的输出变成原来的攻击力 * 攻击时间
此时对全图跑最大流
如果最大流和防御系数之和相等
证明当前时间可以摧毁所有防御机器人
那么就可以缩小答案规模
反之当前输出不够摧毁所有机器人
那么增加答案使输出更接近防御系数之和
考虑精度问题
可以给所有防御系数乘上10000（因为精度限制只有1000）
攻击参数不动
这样跑出来的攻击时间会比原来大10000倍
再将答案除以10000输出即可

Code

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cmath>
#define int long long
#define min(a, b) ({register int aa = a, bb = b; aa > bb ? bb : aa;})
#define inf 0x7fffffff
using namespace std;

inline int read(){
    int x = 0, w = 1;
    char ch = getchar();
    for(; ch > '9' || ch < '0'; ch = getchar()) if(ch == '-') w = -1;
    for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
    return x * w;
}

const int ss = 10010;

struct node{
    int to, nxt, w;
}edge[ss * 20];

int head[ss], tot = 1;
inline void add(register int u, register int v, register int w){
    edge[++tot].to = v;
    edge[tot].nxt = head[u];
    edge[tot].w = w;
    head[u] = tot;
}

int dis[ss], cur[ss];
int n, m, s, t;
bool vis[ss];
queue<int> q;
inline bool spfa(register int s){
    for(register int i = 0; i <= t; i++)
        dis[i] = 0x3f3f3f3f, cur[i] = head[i], vis[i] = 0;
    dis[s] = 0;
    q.push(s);
    while(!q.empty()){
        register int u = q.front();
        q.pop();
        vis[u] = 0;
        for(register int i = head[u]; i; i = edge[i].nxt){
            register int v = edge[i].to;
            if(dis[v] > dis[u] + 1 && edge[i].w){
                dis[v] = dis[u] + 1;
                if(!vis[v]) q.push(v), vis[v] = 1;
            }
        }
    }
    return dis[t] != 0x3f3f3f3f;
}

inline int dfs(register int u, register int flow){
    register int res = 0;
    if(u == t) return flow;
    for(register int i = cur[u]; i; i = edge[i].nxt){
        cur[u] = i;
        register int v = edge[i].to;
        if(dis[v] == dis[u] + 1 && edge[i].w){
            if(res = dfs(v, min(flow, edge[i].w))){
                edge[i].w -= res;
                edge[i ^ 1].w += res;
                return res;
            }
        }
    }
    return 0;
}

inline long long dinic(){
    register long long maxflow = 0;
    register long long minflow = 0;
    while(spfa(s)){
        while(minflow = dfs(s, 0x7fffffff))
            maxflow += minflow;
    }
    return maxflow;
}

long long a[ss], b[ss], sum;
bool map[55][55];
inline bool check(register long long x){
    for(register int j = 1; j <= m; j++)
        b[j] *= x;
    memset(head, 0, sizeof head);
    memset(cur, 0, sizeof cur);
    tot = 1;
    for(register int i = 1; i <= m; i++)
        add(s, i, b[i]), add(i, s, 0);
    for(register int j = 1; j <= n; j++)
        add(j + m, t, a[j]), add(t, j + m, 0);
    for(register int j = 1; j <= m; j++){
        for(register int i = 1; i <= n; i++){
            register bool op = map[j][i];
            if(op == 1){
                add(j, i + m, inf);
                add(i + m, j, 0);
            }
            else continue;
        }
    }
    register long long tmp = dinic();
    register bool flag = 1;
    if(tmp >= sum) flag = 1;
    else flag = 0;
    for(register int j = 1; j <= m; j++)
        b[j] /= x;
    return flag;
}

signed main(){
    n = read(), m = read();
    s = 0, t = n + m + 1;
    for(register int i = 1; i <= n; i++) a[i] = read() * 10000, sum += a[i];
    for(register int j = 1; j <= m; j++) b[j] = read();
    for(register int j = 1; j <= m; j++)
        for(register int i = 1; i <= n; i++)
            map[j][i] = read();
    register long long l = 0, r = 1e9;
    while(l <= r){
        register long long mid = l + r >> 1;
        if(check(mid)) r = mid - 1;
        else l = mid + 1;
    }
    printf("%.6lf\n", (double)l / 10000);
    return 0;
}