题意就是让你算找一个得分最大的子串
代码：
#include <map> #include <set> #include <ctime> #include <stack> #include <queue> #include <cmath> #include <string> #include <vector> #include <cstdio> #include <cctype> #include <cstring> #include <sstream> #include <cstdlib> #include <iostream> #include <algorithm> #pragma comment(linker,"/STACK:102400000,102400000") using namespace std; #define MAX 1000005 #define MAXN 1000005 #define maxnode 205 #define sigma_size 26 #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define lrt rt<<1 #define rrt rt<<1|1 #define middle int m=(r+l)>>1 #define LL long long #define ull unsigned long long #define mem(x,v) memset(x,v,sizeof(x)) #define lowbit(x) (x&-x) #define pii pair<int,int> #define bits(a) __builtin_popcount(a) #define mk make_pair #define limit 10000 //const int prime = 999983; const int INF = 0x3f3f3f3f; const LL INFF = 0x3f3f; const double pi = acos(-1.0); const double inf = 1e18; const double eps = 1e-4; const LL mod = 772002; const ull mx = 133333331; /*****************************************************/ inline void RI(int &x) { char c; while((c=getchar())<'0' || c>'9'); x=c-'0'; while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0'; } /*****************************************************/ const int N = 11111; const int M = 100005; struct Isap{ int tot; int d[N],pre[N],cur[N],gap[N]; int head[N]; struct Edge{ int v,next; int cap,flow; }edge[M*2]; void init(){ mem(head,-1); tot=0; } void add_edge(int a, int b, int c){ edge[tot]=(Edge){b,head[a],c,0}; head[a]=tot++; } void add(int a,int b, int c){ add_edge(a,b,c); add_edge(b,a,0); } void set_d(int t){ queue<int> q; mem(d,-1); mem(gap,0); d[t]=0; q.push(t); while(!q.empty()) { int u=q.front();q.pop(); ++gap[d[u]]; for(int i=head[u];i!=-1;i=edge[i].next) { int v=edge[i].v; if(d[v]==-1) { d[v]=d[u]+1; q.push(v); } } } } int sap(int s,int t,int num) { set_d(t); int ans=0,u=s; int flow=INF; memcpy(cur,head,sizeof(head)); while(d[s]<num){ int &i=cur[u]; for(;i!=-1;i=edge[i].next) { int v=edge[i].v; if(edge[i].cap>edge[i].flow&&d[u]==d[v]+1) { u=v; pre[v]=i; flow=min(flow,edge[i].cap-edge[i].flow); if(u==t){ while(u!=s){ int j=pre[u]; edge[j].flow+=flow; edge[j^1].flow-=flow; u=edge[j^1].v; } ans+=flow; flow=INF; } break; } } if(i==-1) { if(--gap[d[u]]==0) break; int dmin=num-1; cur[u]=head[u]; for(int j=head[u];j!=-1;j=edge[j].next) if(edge[j].cap>edge[j].flow) dmin=min(dmin,d[edge[j].v]); d[u]=dmin+1; ++gap[d[u]]; if(u!=s) u=edge[pre[u]^1].v; } } return ans; } }Sap; /*调用方式： Sap.init(); //建边前调用 Sap.add(u, v, c); //在u->v之间建一条容量为c的边 Sap.sap(s, t, n); //s为源点，t为汇点，n为点的数量*/ char s[105]; int a[15],b[15]; int main(){ //freopen("froggy.in","r",stdin); //freopen("froggy.out","w",stdout); int t,kase=0; cin>>t; while(t--){ int n; kase++; scanf("%d%s",&n,s+1); Sap.init(); for(int i=0;i<10;i++) scanf("%d%d",&a[i],&b[i]); for(int i=0;i<10;i++) Sap.add(n*n+n+i+1,n*n+n+11,b[i]-a[i]); for(int i=1;i<=n;i++){ Sap.add(n*n+i,n*n+n+11,a[s[i]-'0']); Sap.add(n*n+i,n*n+n+(s[i]-'0')+1,INF); } int sum=0; for(int i=1;i<=n;i++){ for(int j=1;j<=n;j++){ int x; scanf("%d",&x); Sap.add(0,(i-1)*n+j,x); sum+=x; //if(i==j) continue; Sap.add((i-1)*n+j,n*n+i,INF); Sap.add((i-1)*n+j,n*n+j,INF); } } printf("Case #%d: %d\n",kase,sum-Sap.sap(0,n*n+n+11,n*n+n+12)); } return 0; }