题目链接:点击这里
解题思路:
如果我们选了(i,j)点对,那么就会获得w[i][j]+w[j][i]的利益.
那么我们在原有的点上增加(n*(n-1))/2个点对点,和10个0~9的字符点.原来的点对应的权值就是改点所对应字符值得ax
0~9字符点的权值就是bx-ax,点对点的权值就是w[i][j]+w[j][i].除了正权连源点S,负权连汇点T之外,(i,j)连j和i,i连对应字符点值得点.
建成图之后,跑最小割即可.
#include<bits/stdc++.h>
#define inf 0x3f3f3f3f
typedef long long ll;
using namespace std;
const int mx = 1e4 + 10;
int n,m,S,T,dep[mx],tot,w[110][110];
int head[mx],cur[mx],ai[20],bi[20];
char str[mx];
struct node
{
int y,nxt;
int c;
}edge[mx*10];
void init()
{
tot = 0;
memset(head,-1,sizeof(head));
}
void AddEdge(int x,int y,int c)
{
edge[tot] = {y,head[x],c};
head[x] = tot++;
edge[tot] = {x,head[y],0};
head[y] = tot++;
}
bool bfs()
{
memset(dep,0,sizeof(dep));
queue <int> que;
que.push(S);
dep[S] = 1;
while(!que.empty())
{
int no = que.front();
que.pop();
for(int i=head[no];~i;i=edge[i].nxt)
{
int u = edge[i].y;
if(!dep[u]&&edge[i].c){
dep[u] = dep[no] + 1;
que.push(u);
}
}
}
return dep[T];
}
int dfs(int x,int flow)
{
if(x==T||!flow) return flow;
int used = 0;
for(int i=head[x];~i;i=edge[i].nxt)
{
int u = edge[i].y;
if(dep[x]+1==dep[u]){
int w = dfs(u,min(edge[i].c,flow-used));
edge[i].c -= w;
edge[i^1].c += w;
used += w;
if(used==flow) return flow;
}
}
if(!used) dep[x] = 0;
return used;
}
int maxflow()
{
int ans = 0;
while(bfs()){
//copy(head,head+T+1,cur);
ans += dfs(S,inf);
}
return ans;
}
int main()
{
int t,a,k,ca = 1;
scanf("%d",&t);
while(t--){
int ret = 0;
init();
scanf("%d",&n);
scanf("%s",str+1);
int cnt = n+10;
for(int i=0;i<10;i++) scanf("%d%d",ai+i,bi+i);
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++)
scanf("%d",w[i]+j);
}
for(int i=1;i<=n;i++){
for(int j=i+1;j<=n;j++){
ret += w[i][j] + w[j][i];
AddEdge(S,++cnt,w[i][j]+w[j][i]);
AddEdge(cnt,i,inf);
AddEdge(cnt,j,inf);
}
}
for(int i=1;i<=n;i++) AddEdge(i,n+1+str[i]-'0',inf);
T = cnt + 1;
for(int i=1;i<=n;i++) AddEdge(i,T,ai[str[i]-'0']);
for(int i=1;i<=10;i++) AddEdge(n+i,T,bi[i-1]-ai[i-1]);
printf("Case #%d: %d\n",ca++,ret-maxflow());
}
return 0;
}