POJ3613

POJ3613

---

题意：n条边构成的最短路

做法：倍增floyd

#include <cstdio>
#include <cstring>
#include <algorithm>
#define rep(i,a,b) for(int i=a;i<=b;++i)
typedef long long ll;
const int N = 207;
const int inf = 0x3f3f3f3f;
using namespace std;
int n,cnt,id[1007],T,S,E;
int dis[N][N][25],d[N][N],tmp[N][N];// dis[i][j][k] : i -> j 经过 2^k 条边的最短路

int main() {
    scanf("%d%d%d%d",&n,&T,&S,&E);
    rep(i,1,200)rep(j,1,200)rep(k,0,22) dis[i][j][k] = d[i][j] = inf;
    rep(i,1,T) {int x,y,z;
        scanf("%d %d %d",&z,&x,&y);
        if(!id[x]) id[x] = ++cnt;
        if(!id[y]) id[y] = ++cnt;
        dis[id[x]][id[y]][0] = dis[id[y]][id[x]][0] = min(z,dis[id[y]][id[x]][0]);
    }
    rep(s,1,20)rep(k,1,cnt)rep(i,1,cnt)rep(j,1,cnt){
        dis[i][j][s] = min(dis[i][j][s],dis[i][k][s-1]+dis[k][j][s-1]);
    }
    rep(i,1,cnt) d[i][i] = 0;
    rep(s,0,20)if(n&(1<<s)) {
        rep(i,1,cnt)rep(j,1,cnt) tmp[i][j]=inf;
        rep(k,1,cnt)rep(i,1,cnt)rep(j,1,cnt) {
            tmp[i][j] = min(tmp[i][j],d[i][k]+dis[k][j][s]);
        }
        rep(i,1,cnt)rep(j,1,cnt)d[i][j]=tmp[i][j];
    }
    printf("%d\n",d[id[S]][id[E]]);
    return 0;
}