A few days before the first title of the pigeon
T1 mine
Simple $ dp $, meaning you can divert attention.
My definition of too much trouble, but I refer to other chiefs of blog attention.
Defined $ f_ {i, j, k} $ indicates the current position $ i $, the current element is $ j $, the $ k $ is an element of the program number. i dimension you can roll up.
Nothing to say, little trouble is $? $ And $ 1 $, and the other to simulate transfer it.
Brother Fucai.
1 #include<cstdio> 2 #include<string> 3 #include<iostream> 4 #include<cstring> 5 #define LL long long 6 #define HZOI std 7 using namespace HZOI; 8 const int mod=1e9+7; 9 LL ans; 10 int n; 11 char s[1000003],a[1000003]; 12 LL dp[1000003][5][5]; 13 inline void Mod(LL &a) {while (a>=mod) a-=mod;} 14 int main() 15 { 16 char cc=getchar(); 17 while (cc!='0' and cc!='1' and cc!='2' and cc!='*' and cc!='?') cc=getchar(); 18 while (cc=='0' or cc=='1' or cc=='2' or cc=='*' or cc=='?') 19 { 20 if (cc=='?') a[++n]=4; 21 else if (cc=='*') a[++n]=3; 22 else a[++n]=cc-48; 23 cc=getchar(); 24 } 25 if (a[1]!=4 and a[2]!=4) dp[2][a[2]][a[1]]=1; 26 if (a[1]==4 and a[2]!=4) 27 { 28 if (!a[2]) 29 dp[2][a[2]][0]=1; 30 if (a[2]==1) 31 dp[2][a[2]][0]=dp[2][a[2]][3]=1; 32 if (a[2]==2) 33 dp[2][a[2]][3]=1; 34 if (a[2]==3) 35 dp[2][a[2]][3]=dp[2][a[2]][1]=1; 36 } 37 if (a[1]!=4 and a[2]==4) 38 { 39 if (!a[1]) 40 dp[2][0][a[1]]=dp[2][1][a[1]]=1; 41 if (a[1]==1) 42 dp[2][3][a[1]]=1; 43 if (a[1]==2) return puts("0"),0; 44 if (a[1]==3) 45 dp[2][1][a[1]]=dp[2][2][a[1]]=dp[2][3][a[1]]=1; 46 } 47 if (a[1]==4 and a[2]==4) 48 { 49 dp[2][0][0]=dp[2][1][0]=dp[2][3][1]=dp[2][1][3]=dp[2][2][3]=dp[2][3][3]=1; 50 } 51 for (int i=3; i<=n; ++i) 52 { 53 if (!a[i]) 54 for (int j=0; j<3; ++j) 55 { 56 if (!j) 57 for (int k=0; k<2; ++k) 58 Mod(dp[i][0][j]+=dp[i-1][j][k]); 59 if (j==1) 60 Mod(dp[i][0][j]+=dp[i-1][j][3]); 61 } 62 if (a[i]==1) 63 for (int j=0; j<4; ++j) 64 { 65 if (!j) 66 for (int k=0; k<2; ++k) 67 Mod(dp[i][1][j]+=dp[i-1][j][k]); 68 if (j==1) 69 Mod(dp[i][1][j]+=dp[i-1][j][3]); 70 if (j==3) 71 for (int k=1; k<4; ++k) 72 Mod(dp[i][1][j]+=dp[i-1][j][k]); 73 } 74 if (a[i]==2) 75 for (int j=1; j<4; ++j) 76 Mod(dp[i][2][3]+=dp[i-1][3][j]); 77 if (a[i]==3) 78 for (int j=1; j<4; ++j) 79 { 80 if (j==1) 81 for (int k=0; k<2; ++k) 82 Mod(dp[i][3][j]+=dp[i-1][j][k]); 83 if (j==2) 84 Mod(dp[i][3][j]+=dp[i-1][j][3]); 85 if (j==3) 86 for (int k=1; k<4; ++k) 87 Mod(dp[i][3][j]+=dp[i-1][j][k]); 88 } 89 if (a[i]==4) 90 for (int g=0; g<4; ++g) 91 { 92 if (!g) 93 for (int j=0; j<3; ++j) 94 { 95 if (!j) 96 for (int k=0; k<2; ++k) 97 Mod(dp[i][0][j]+=dp[i-1][j][k]); 98 if (j==1) 99 Mod(dp[i][0][j]+=dp[i-1][j][3]); 100 } 101 if (g==1) 102 for (int j=0; j<4; ++j) 103 { 104 if (!j) 105 for (int k=0; k<2; ++k) 106 Mod(dp[i][1][j]+=dp[i-1][j][k]); 107 if (j==1) 108 Mod(dp[i][1][j]+=dp[i-1][j][3]); 109 if (j==3) 110 for (int k=1; k<4; ++k) 111 Mod(dp[i][1][j]+=dp[i-1][j][k]); 112 } 113 if (g==2) 114 for (int j=1; j<4; ++j) 115 Mod(dp[i][2][3]+=dp[i-1][3][j]); 116 if (g==3) 117 for (int j=1; j<4; ++j) 118 { 119 if (j==1) 120 for (int k=0; k<2; ++k) 121 Mod(dp[i][3][j]+=dp[i-1][j][k]); 122 if (j==2) 123 Mod(dp[i][3][j]+=dp[i-1][j][3]); 124 if (j==3) 125 for (int k=1; k<4; ++k) 126 Mod(dp[i][3][j]+=dp[i-1][j][k]); 127 } 128 } 129 } 130 Mod(ans+=dp[n][0][0]+dp[n][0][1]+dp[n][1][3]+dp[n][3][1]+dp[n][3][2]+dp[n][3][3]); 131 printf("%lld\n",(ans+mod)%mod); 132 }
T2 water
When there is no in-depth examination of understanding the problem, resulting in Mongolia to $ 20 $ points.
In fact, the idea is not difficult, but I did not expect when the examination.
This question is a main idea, from the beginning to the inside border sweep with state borders update the internal state, you can find a minimum flow out of the mouth, which is the key to solving the problem.
We can consider $ SPFA $ traverse the entire map, and then continue to update its value with the dis answer, be careful not to be considered the starting point of $ dis $, should also consider the value end of $ a $, whichever is $ max $, because I make sure the answer to the situation "back" does not appear.
Brother Fucai.
1 #include<cstdio> 2 #include<cstring> 3 #define LL long long 4 #define HZOI std 5 using namespace HZOI; 6 int n,m; 7 LL head,tail,a[303][303],dis[303][303],vis[303][303],quei[1000003],quej[1000003]; 8 int dx[4]={0,0,1,-1},dy[4]={1,-1,0,0}; 9 inline void Bfs(); 10 inline LL read(); 11 inline LL max(LL a,LL b) {return a>b?a:b;} 12 inline LL min(LL a,LL b) {return a<b?a:b;} 13 int main() 14 { 15 n=read(),m=read(); 16 for (int i=1; i<=n; ++i) 17 for (int j=1; j<=m; ++j) 18 a[i][j]=read(); 19 memset(dis,0x3f,sizeof(dis)); 20 head=tail=0; 21 for (int i=1; i<=n; ++i) 22 { 23 dis[i][m]=max(0,a[i][m]); 24 dis[i][1]=max(0,a[i][1]); 25 if (!vis[i][m]) quei[++tail]=i, quej[tail]=m, vis[i][m]=1; 26 if (!vis[i][1]) quei[++tail]=i, quej[tail]=1, vis[i][1]=1; 27 } 28 for (int i=1; i<=m; ++i) 29 { 30 dis[n][i]=max(0,a[n][i]); 31 dis[1][i]=max(0,a[1][i]); 32 if (!vis[n][i]) quei[++tail]=n, quej[tail]=i, vis[n][i]=1; 33 if (!vis[1][i]) quei[++tail]=1, quej[tail]=i, vis[1][i]=1; 34 } 35 Bfs(); 36 for (int i=1; i<=n; ++i) 37 { 38 for (int j=1; j<=m; ++j) 39 printf("%lld ",dis[i][j]-a[i][j]); 40 puts(""); 41 } 42 } 43 inline void Bfs() 44 { 45 while (head^tail) 46 { 47 int xi=quei[++head],xj=quej[head]; 48 vis[xi][xj]=0; 49 for (int i=0; i<4; ++i) 50 { 51 int toi=xi+dx[i],toj=xj+dy[i]; 52 if (toi>0 and toi<=n and toj>0 and toj<=m) 53 { 54 int dtc=max(a[toi][toj],dis[xi][xj]); 55 if (dis[toi][toj]>dtc) 56 { 57 dis[toi][toj]=dtc; 58 if (!vis[toi][toj]) quei[++tail]=toi,quej[tail]=toj,vis[toi][toj]=1; 59 } 60 } 61 } 62 } 63 } 64 inline LL read() 65 { 66 LL nn=0,ff=1; char cc=getchar(); 67 while (cc<'0' or cc>'9') {if (cc=='-') ff=-1; cc=getchar();} 68 while (cc>='0' and cc<='9') nn=(nn<<3)+(nn<<1)+(cc^48),cc=getchar(); 69 return nn*ff; 70 }