9.1 联合作战战果

1.处理内容：

树链剖分4题

最大流模板1题

线性DP3题

bellman-ford判负环1题

2.树剖

直接甩链接

3.最大流模板 略过

4.线性DP

水题1：目测是USACO2009Open的滑雪课

f[i][j]为i时能力j的最大滑雪次数

第一种转移：我什么都不做这是坠吼的——用f[i][j]更新f[i+1][j]

第二种转移：上课，当然是当前时间且值更高——用f[i][j]更新f[i+l][a]

第三种转移：滑雪，当前能力最大值，稍微要点预处理——用f[i][j]+1更新f[i+d][j]

局部完结撒花

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<ctime>#include<cmath>#include<algorithm>#include<cctype>#include<iomanip>using namespace std;inline int read(){int i=0,f=1;char ch;for(ch=getchar();!isdigit(ch);ch=getchar())if(ch=='-') f=-1;for(;isdigit(ch);ch=getchar())i=(i<<1)+(i<<3)+(ch^48);return i*f;}int buf[1024];inline void write(int x){if(!x){putchar('0');return ;}if(x<0){putchar('-');x=-x;}while(x){buf[++buf[0]]=x%10,x/=10;}while(buf[0]) putchar(buf[buf[0]--]+48);return ;}#define stan1 111#define stan2 11111int t,s,n,m,l,a,x,y,nxt[stan1],first[stan2],tme[stan1],val[stan1],bst[stan1],f[stan2][stan1],ans;signed main(){t=read();s=read();n=read();for(int i=1;i<=s;++i){m=read();tme[i]=read();val[i]=read();nxt[i]=first[m];first[m]=i;}for(int i=1;i<=100;++i) bst[i]=999999;for(int i=1;i<=n;++i){x=read();y=read();if(bst[x]>y) bst[x]=y;}for(int i=1;i<100;++i)if(bst[i]<bst[i+1])bst[i+1]=bst[i];memset(f,128,sizeof(f));f[0][1]=0;for(int i=0;i<t;++i)for(int j=1;j<=100;++j)if(f[i][j]>=0){if(f[i+1][j]<f[i][j])f[i+1][j]=f[i][j];if(i+bst[j]<=t)f[i+bst[j]][j]=max(f[i+bst[j]][j],f[i][j]+1);for(int p=first[i];p;p=nxt[p]){if(i+tme[p]<=t&&val[p]>j)f[i+tme[p]][val[p]]=max(f[i+tme[p]][val[p]],f[i][j]);}}for(int j=1;j<=100;++j)if(f[t][j]>ans)ans=f[t][j];write(ans);return 0;}

水题2：Bbj

A 和 B 玩游戏。每回合，每个人需要作出下列决策之一：

1. 充能：增加一点能量值

2. 攻击：消耗一点能量值，攻击对方

3. 防御：可以抵挡对方的攻击

两个人同时攻击时视作抵消。某人被攻击时若在充能则该人输去。一开始每个人的能量均为零。已知游戏经过了 n 回合后仍未分出胜负，求有多少种可能的游戏

局面。答案模 1e9 + 7。

因为n<=200 所以你用dfs直接爆搜肯定药丸

定义f[x][y][z]表示x回合甲能量为y乙能量为z

然后n^3再次局部完结撒花

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<ctime>#include<cmath>#include<algorithm>#include<cctype>#include<iomanip>#define mod 1000000007using namespace std;inline int read(){int i=0,f=1;char ch;for(ch=getchar();!isdigit(ch);ch=getchar())if(ch=='-') f=-1;for(;isdigit(ch);ch=getchar())i=(i<<1)+(i<<3)+(ch^48);return i*f;}int buf[1024];inline void write(long long x){if(!x){putchar('0');return ;}if(x<0){putchar('-');x=-x;}while(x){buf[++buf[0]]=x%10,x/=10;}while(buf[0]) putchar(buf[buf[0]--]+48);return ;}int n,f[222][222][222];int dfs(int x,int y,int z){if(y<0||z<0) return 0;if(x>n) return 1; if(f[x][y][z]!=-1) return f[x][y][z];f[x][y][z]=0;for(int i=-1;i<=1;++i)for(int j=-1;j<=1;++j)if((i==1&&j==-1)||(j==1&&i==-1)) continue;else f[x][y][z]=(f[x][y][z]+dfs(x+1,y+i,z+j))%mod;return f[x][y][z];}signed main(){memset(f,-1,sizeof(f));n=read();write(dfs(1,0,0));return 0;}

难题：地精部落

容我想想，卒

临死决定甩个链接

http://blog.csdn.net/mrazer/article/details/53426415?locationNum=12&fps=1

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<ctime>#include<cmath>#include<algorithm>#include<cctype>#include<iomanip>using namespace std;inline int read(){int i=0,f=1;char ch;for(ch=getchar();!isdigit(ch);ch=getchar())if(ch=='-') f=-1;for(;isdigit(ch);ch=getchar())i=(i<<1)+(i<<3)+(ch^48);return i*f;}int buf[1024];inline void write(int x){if(!x){putchar('0');return ;}if(x<0){putchar('-');x=-x;}while(x){buf[++buf[0]]=x%10,x/=10;}while(buf[0]) putchar(buf[buf[0]--]+48);return ;}int n,p,f[2][5555];signed main(){n=read();p=read();f[1][1]=1;for(int i=2;i<=n;++i)for(int j=1;j<=i;++j)f[i&1][j]=(f[i&1][j-1]+f[(i&1)^1][i-j])%p;write((f[n&1][n]<<1)%p);return 0;}

4.bellman-ford 略

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<ctime>#include<cmath>#include<algorithm>#include<cctype>#include<iomanip>using namespace std;inline int read(){int i=0,f=1;char ch;for(ch=getchar();!isdigit(ch);ch=getchar())if(ch=='-') f=-1;for(;isdigit(ch);ch=getchar())i=(i<<1)+(i<<3)+(ch^48);return i*f;}int buf[1024];inline void write(int x){if(!x){putchar('0');return ;}if(x<0){putchar('-');x=-x;}while(x){buf[++buf[0]]=x%10,x/=10;}while(buf[0]) putchar(buf[buf[0]--]+48);return ;}int x,cnt[555],tot,n,m,w,s,e,t,to[555],from[5555],first[555],nxt[5555],goal[5555],dis[5555],exi[555];void addedge(int a,int b,int c){++tot;nxt[tot]=first[a];first[a]=tot;from[tot]=a;goal[tot]=b;dis[tot]=c;return ;}bool check(){for(int i=2;i<=n;++i)to[i]=999999999;for(int i=1;i<n;++i){int tag=0;for(int p=1;p<=tot;++p)if(to[goal[p]]>to[from[p]]+dis[p]){to[goal[p]]=to[from[p]]+dis[p];tag=1;}if(!tag) break;}for(int p=1;p<=tot;++p)if(to[goal[p]]>to[from[p]]+dis[p])return true;return false;}signed main(){x=read();while(x--){tot=0;memset(cnt,0,sizeof(cnt));memset(first,0,sizeof(first));n=read();m=read();w=read();for(int i=1;i<=m;++i){s=read();e=read();t=read();addedge(s,e,t);addedge(e,s,t);}for(int i=1;i<=w;++i){s=read();e=read();t=read();addedge(s,e,-1*t);}if(check()) puts("YES");else puts("NO");}return 0;}