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/* dp滚动数组 第一次用滚动数组 感觉挺好 之前用二维double型数组 MLE 在运行结果里看到有人用4080K过了（限制4M） 神人也。。。 */ #define LOCAL #include<iostream> #include<cstdio> #include<cmath> #include<cstring> #include<cstdlib> #include<iomanip> #include<string> #include<algorithm> #include<ctime> #include<stack> #include<queue> #include<vector> #define N 1005 double dp1[N],dp2[N]; bool cross[N][N]; double minthree(double a,double b,double c) {if(a<=b&&a<=c)return a; else if(b<=a&&b<=c)return b; else if(c<=a&&c<=b)return c;} double mintwo(double a,double b) {if(a<b)return a;else return b;} using namespace std; int main() { #ifdef LOCAL freopen("input.txt","r",stdin); freopen("output.txt","w",stdout); #endif int n,m,ncross,i,j,x,y; double diag=100*sqrt((double)2); cin>>m>>n>>ncross;m++;n++; memset(dp2,0,sizeof(dp2)); memset(cross,0,sizeof(cross)); for(i=0;i<ncross;i++) {cin>>x>>y;cross[x][y]=true;} dp1[1]=0; for(i=2;i<=n;i++) {dp1[i]=dp1[i-1]+100;} for(i=2;i<=m;i++) { for(j=1;j<=n;j++) { if(j==1) {dp2[j]=dp1[j]+100;} else if(cross[i-1][j-1]) {dp2[j]=minthree(dp1[j-1]+diag,dp1[j]+100,dp2[j-1]+100);} else {dp2[j]=mintwo(dp1[j]+100,dp2[j-1]+100);} } memcpy(dp1,dp2,sizeof(double)*N); memset(dp2,0,sizeof(dp2)); } cout<<(int)(dp1[n]+0.5)<<endl; return 0; }