字符串终于告一段落了！

题目链接: (bzoj) https://www.lydsy.com/JudgeOnline/problem.php?id=3993

(luogu) https://www.luogu.org/problemnew/show/P3324

网络流从最水的开始做。。。

题解: 二分答案ans, 然后可以得到每个攻击者在ans时间内最多产生的总伤害，从起点往攻击者连边容量为此值，从每个攻击者往能攻击的防御者连边容量为\(+\inf\), 从每个防御者往终点连边边权为其装甲值，求最大流，判断其是否等于装甲值之和即可。

时间复杂度\(O(MaxFlow(n,n^2))\)

代码

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;const int N = 102;
const int M = 2600;
const double INF = 1e7;
const double EPS = 1e-8;namespace MaxFlow
{struct Edge{int v,nxt,rev; double w;} e[(M<<1)+3];int fe[N+3];int te[N+3];int dep[N+3];int que[N+3];int n,en;void clear(){for(int i=1; i<=en; i++){e[i].v = e[i].nxt = e[i].rev = 0; e[i].w = 0.0;}for(int i=1; i<=n; i++) fe[i] = 0;n = en = 0;}void addedge(int u,int v,double w){en++; e[en].v = v; e[en].w = w;e[en].nxt = fe[u]; fe[u] = en; e[en].rev = en+1;en++; e[en].v = u; e[en].w = 0;e[en].nxt = fe[v]; fe[v] = en; e[en].rev = en-1;}bool bfs(){for(int i=1; i<=n; i++) dep[i] = 0;int head = 1,tail = 1; que[tail] = 1; dep[1] = 1;while(head<=tail){int u = que[head]; head++;for(int i=fe[u]; i; i=e[i].nxt){if(dep[e[i].v]==0 && e[i].w>EPS){dep[e[i].v] = dep[u]+1;tail++; que[tail] = e[i].v;}}}return dep[2]!=0;}double dfs(int u,double cur){if(u==2) return cur;double rst = cur;for(int i=te[u]; i; i=e[i].nxt){if(dep[e[i].v]==dep[u]+1 && e[i].w>0 && rst>0){double flow = dfs(e[i].v,min(rst,e[i].w));if(flow>EPS){rst -= flow; e[i].w -= flow; e[e[i].rev].w += flow;if(e[i].w>EPS) te[u] = i;if(rst==0) return cur;}}}if(abs(cur-rst)<EPS) dep[u] = 0;return cur-rst;}double dinic(int _n){n = _n;double ret = 0.0;while(bfs()){for(int i=1; i<=n; i++) te[i] = fe[i];ret += dfs(1,INF);}return ret;}
}int a[N+3],b[N+3];
int c[N+3][N+3];
int n,m;int main()
{scanf("%d%d",&n,&m); double std = 0.0;for(int i=1; i<=n; i++) {scanf("%d",&a[i]); std += (double)a[i];}for(int i=1; i<=m; i++) scanf("%d",&b[i]);for(int i=1; i<=m; i++){for(int j=1; j<=n; j++) scanf("%d",&c[i][j]);}double left = 0.0,right = INF;while(right-left>1e-5){double mid = (left+right)*0.5;MaxFlow::clear();for(int i=1; i<=m; i++){MaxFlow::addedge(1,i+2,mid*(double)b[i]);}for(int i=1; i<=n; i++){MaxFlow::addedge(i+m+2,2,(double)a[i]);}for(int i=1; i<=m; i++){for(int j=1; j<=n; j++){if(c[i][j]==1){MaxFlow::addedge(i+2,j+m+2,INF);}}}double ans = MaxFlow::dinic(m+n+2);if(abs(ans-std)<EPS){right = mid;}else{left = mid;}}printf("%.6lf\n",left);return 0;
}